UC-NRLF 


$B    53D    7ET 


LIBRARY 

OF  THE 

University  of  California. 

Mrs.  SARAH  P.  WALSWORTH. 

Recei^d  October,  i8g4. 
z/Iccessions  No.Q^y'/3*5~^     Class  No, 


3< 


V'\ '■•••-  \jM,  •  -     • 


"% 


ELEMENTS 


OF 


GEOMETRY  AND  TRIGONOMETRY. 


TRANSLATED  FRO^  THE  FRENCH  OF 

A.  M.  LE'GENDRE, 

BY    DAVID    BREWSTER,   LL.   D. 


f 


REVISED  AND  ADAPTED  TO  THE  COURSE  OF.  MATHEMATICAL  INSTRUCTION* 
IN  THE  UNITED  STATES, 

BY    CHARLES  DAVIES, 

AUTHOR  OF  MENTAL  AND  PRACTICAL   ARITHMETIC,  ELEMENTS  OF  SURVEYING, 
ELEMENTS    OP    DESCRIPTIVE    AND    OF    ANALYTICAL    GEOMETRY, 
ELEMENTS  OF  DIFFERENTIAL  AND  INTEGRAL  CALCULUS, 
AND  SHADES  SHADOWS  AND  PERSPECTIVE. 


PHILADELPHIA : 


PUBLISHED   BY   A.S.BARNES  AND   CO. 
21  Minor-street. 


DAVIES'  ^1i^ 

COpRSE   OF   MATHEMATICS. 


DA  VIES'  FIRST  LESSONS  IN  ARITHMETIC, 

DESIGNED  FOR  BEGINNERS. 


DA  VIES'  ARITHMETIC, 

DESIGNED  FOR  THE  USE  OF  ACADEMIES  AND  SCHOOCS. 


KEY  TO  DAVIES'  ARITHMETIC. 


DA  VIES'  FIRST  LESSONS  IN  ALGEBRA; 

Being  an  introduction  to  the  Science,  and  forming  a  connecting  link  between 

Arithmetic  and  Algebra. 


DAVIES'  ELEMENTS  OF  GEOMETRY. 

This  work  embraces  the  elementary  principles  of  Geometry.     The  reasoning  is  plain 

and  concise,  but  at  the  same  time  strictly  rigorous. 


DAVIES'  PRACTICAL  GEOMETRY, 

Embracing  the  facts  of  Geometry,  with  applications  in  Artificer's  Work, 

Mensuration  and  Mechanical  Philosophy. 


DAVIES'  BOURDON'S  ALGEBRA, 
Bemg  an  abridgment  of  the  work  of  M.  Bourdon,  with  the  addition  of  practical 

examples. 


DAVIES'  LEGENDRE'S  GEOMETRY  and  TRIGONOMETRY, 

Being  an  abridgment  of  the  work  of  M.  Legendre,  with  the  addition  of  a  Treatise 

on  Mensuration  of  Planes  and  Solids,  and  a  Table  of  Logarithms  and 

Logarithmic  Sines. 


DAVIES'  SURVEYING, 

With  a  description  and  plates  of,  the  Theodolite,-  Compass,  Plane-Table,  and 

Level — also,  Maps  of  the  Topographical  Signs  adopted  by  the  Engineer 

Department — an   explanation    of    the  method  of    surveying 

the  Public  Lands,  and  an  Elementary  Treatise  on 

Navigation. 


DAVIES'  ANALYTICAL  GEOMETRY, 

Embracing  the  Equations  of  the  Point  and  Straight  Line — of  the  Conic 

Sections — of  the  Line  and  Pla^e  in  Space — also,  the  discussion  of  the 

General  Equation  of  the  second  degree,  and  of  Surfaces 

OF  the  second  order. 


DAVIES'  DESCRIPTIVE  GEOMETRY, 
With  its  application  to    Spherical  Projections. 

DAVIES'  SHADOWS  and  LINEAR  PERSPEPTIVE. 


DAVIES'  DIFFERENTIAL  and  INTEGRAL  CALCULUS. 


Entered  according  to  the  Act  of  Congress,  in  the  year  1834, 

By    CHARLES    DAVIES, 

in  the  Clerk's  Office  of  the  District  Court  of  the  United  States,  for  the  Southern  District  of  New  York. 


PREFACE 

TO  THE  AMERICAN  EDITION. 


The  Editor,  in  offering  to  the  public  Dr.  Brewster'a 
translation  of  Legendre's  Geometry  under  its  present 
^     form,  is  fully  impressed  with  the  responsibility  he 
assumes  in  making  alterations  in  a  work  of  such  de- 
served celebrity. 

In  the  original  work,  as  well  as*  in  the  translations 
'  of  Dr.  Brewster  and  Pr^essor  Farrar,  the  proposi- 
tions  are  not  enunciated  in  general  terms,  but  with 
reference  to,  and  by  the  aid  of,  the  particular  diagrams 
used  for  the  demonstrations.  It  is  believed  that  this 
departure  from  the  method  of  Euclid  has  been  gene- 
rally regretted.  The  propositions  of  Geometry  are 
general  truths,  and  as  such,  should  be  stated  in  gene- 
ral terms,  and  without  reference  to  particular  figures. 
The  method  of  enunciating  them  by  the  aid  of  particu- 
lar diagrams  seems  to  have  been  adopted  to  avoid  the 
difficulty  which  beginners  experience  in  comprehend- 
ing abstract  propositions.  But  in  avoiding  this  diffi- 
culty, and  thus  lessening,  at  first,  the  intellectual 
labour,  the  faculty  of  abstraction,  which  it  is  one  of 
the  primary  objects  of  the  study  of  Geometry  to 
strengthen,  remains,  to  a  certain  extent,  unimproved. 


iv  .  PREFACE.  , 

Besides  the  alterations  in  the  enunciation  of  the 
propositions,  others  of  considerable  importance  have 
also  been  made  in  the  present  edition.  The  propo- 
sition in  Book  V.,  which  proves  that  a  polygon  and  '# 
circle  may  be  made  to  coincide  so  nearly,  as  to  differ 
from  each  other  by  less  than  any  assignable  quantity, 
has  been  taken  from  the  Edinburgh  Encyclopedia. 
It  is  proved  in  the  corollaries  that  a  polygon  of  ap  •*. 
infinite  number  of  sides  becomes  a  circle,  and  this 
prmciple  is  made  the  basis  of  several  important  de- 
monstrations in  Book  VIIL  4- 

*  •>♦•' 

Book  Il.jOn  Ratios  and  Proportions,  has  been  partly 
adopted  from  the  Encyclopedia  Metropolitana,  and 
will,  it  is  believed,  supply  a  deficiency  in  the  original 
work. 

Very  considerable  alterations  have  also  been  made 
in  the  manner  of  treating  the  subjects  of  Plane  and 
Spherical  Trigonometry.  It  has  also  been  thought 
best  to  publish  with  the  present  edition  a  table  of 
logarithms  and  logarithmic  sines,  and  to  apply  the 
principles  of  geometry  to  the  mensuration  of  sur- 
faces and  solids. 
Military  Academy, 

West  Point,  March,  1834. 


1 


CONTENTS 


The  principles, 
Ratios  and  Proportions, 


BOOK  I. 
BOOK  II. 


BOOK  III. 

The  Circle  and  the  Measurement  of  Angles, 
Problems  relating  to  the  First  and  Third  Books, 

BOOK  IV. 
The  Proportions  of  Figures  and  the  Measurement  of  Areas, 
Problems  relating  to  the  Fourth  Book, 

BOOK  V. 
Regular  Polygons  and  the  Measurement  of  the  Circle, 

BOOK  VI. 
Planes  and  Solid  Angles,  -         - 

BOOK  VII. 
Polyedrons, 

BOOK  vm. 

The  three  round  bodies,   -        .        -        -        - 

BOOK  IX. 
Of  Spherical  Triangles  and  Spherical  Polygons,  - 

APPENDIX. 
The  regular  Polyedrons,  -   .     - 


84       ] 

41 
67 

68 
98 

10? 

126 

142 

166 

186 

209 


vi  CONTENTS. 

PLANE  TRIGONOMETRY, 

Division  of  the  Circumference,  -         -         -         -         -  207 

General  Ideas  relating  to  the  Trigonometrical  Lines,    -         -  208 
Theorems  and  Formulas  relating  to  the  Sines,  Cosines,  Tan- 

gents,  &c.         -         .  -         '        '        -         -  215 

Construction  and  Description  of  the  Tables,  -        -        -  223 

Description  of  Table  of  Logarithms,   -         -         -         -         -  224 

Description  of  Table  of  Logarithmic  Sines,  ...  228 

Principles  for  the  Solution  of  Rectilineal  Triangles,       -         -  231 

Solution  of  Rectilineal  Triangles  by  Logarithms,  -         -  235 

Solution  of  Right  angled  Triangles,  ...         -  237 

Solution  of  Triangles  in  general, 238 

SPHERICAL  TRIGONOMETRY. 

First  principles,       --------  246 

Napier's  Circular  Parts, 252 

Solution  of  Right  angled  Spherical  Triangles  by  Logarithms,  255 

Qiiadrantal  Triangles,      -------  257 

Solution  of  Oblique  angled  Triangles  by  Logaritlims,  -        -  250 

MENSURATION. 

Mensuration  of  Surfaces,  -        -        -        -        -        -  274 

Mensuration  of  Solids,     -------  285 


Alf   INDEX 

SHOWING  THE  PROPOSITIONS    OF   LEGENDRE    WHICH    CORRESPOND  TO 
THE  PRINCIPAL  PROPOSITIONS  OF  THE  FIRST  SIX  BOOKS  OF  EUCLID. 


Euclid. 

Legendre. 

Euclid. 

Legendre. 

Euclid. 

Legendre.   1 

Book  I. 

Book  I. 

Cor.  2.  of  3*2 

Prop.   27 

Prop.   26 

Prop.    15i 

33 
34 

30 

28 

28 
29 

5: 
5 

Prop.    4 

Prop.    5 

5 
Cor.  of  5 

11 

Cor.  of  n 

Book  IV. 

S  Cor.  2.  >iQ 
}    ^3.  r 

6 

12 

35 

1 

31 

8 
13 

10 

1 

36 
37 

1 

Cor.  2.  of  2 

Book  IV. 

14 

3 

38 

Cor.  2.  of  2 

35 

28 

15 

4 

4 

2 

36 

30' 

Cor.  1.5, 5 

&  2.  )  ^^ 

16 

17 

Sch.  of  4 

5  Cor.  of 25 
)     25 

47 

11 

8 

5  Cor.  1.  of  4 
\  Cor.  of  6 

Book  VI. 

Book  II. 

1 

4 

18 

13 

12 

13 

19 
20 
21 
24 

13 

13 

12 

2 

5     15! 

16l 

171 
18| 

7 
8 
9 

Book  111. 

Book  III. 

3 

4 

Prop.    3 

Prop.    6 

25 

9 

10 

Cor.  of   7 

5 

19 

26 

6 

11 

Cor.  of  14 

6 

20! 

27 

Cor.  1.  of  19 

12 

Cor.  of  14 

8 

22 

28 

Cor.  2.  of  19 

14 

8 

14 

)       25 
I  Cor.  of  15 

29 

\  Cor.  2.  & 
I    4.  of  20 

15 

2 

15 

18 

9 

19 

25! 

30 

22 

20 

18 

S                26! 

I                27| 

'Cor.l.of32 

26 

21 

Cor.  of  18 

20 

i 


Digitized.'by  the  Internet  Archive 

in  2008  with^unding  from 

IVIicrosoft  Corporation 


http://www.archive.org/details/elementsofgeometOOIegerich' 


ELEMENTS  OF  GEOMETRY. 

BOOK  I. 

THE  I^INCIPLES. 

Definitions. 

I.  Geometry  is  the  science  which  has  for  it|  object  the 
ij^'isurement  of  extension. 

^^Ixtension  has  tiiree  dimensions,  length,  breadth,  and  height, 
^v  thickness. 

*3.  A  line  i§  length  without  breadth,  or  thickness.  ^ 

The  extremities  of  a  line  arc  called  points :  a  point,. there- 
fyre,  has  neither  length,  breadth,  nor  thickness,  but  position 
only. 

3.  A  straight  line  is  the  shortest  distance  from  one  point  to 
another. 

4.  Every  line  which  is  not  straight,  or  composed  of  straight 
lines,  is  a  curved  line. 

Thus,  AB  is  a  straight  line ;  ACDB  is  a 
broken  line,  or  one   composed  o|  straight  A.< 
lines ;  and  AEB  is  a  curved  line. 

The  word  line,  when  used  alone,  will  designate  a  scraight 
line  ;  and  the  word  curve,  a  curved  line. 

5.  A  surface  is  that  which  has  length  and  breadth,  without 
height  or  thickness. 

0.  A  plane  is  a  surface,  in  which,  if  two  points  be  assumed 
at  pleasure,  and  connected  by  a  straight  line,  that  line  will  lie 
wholly  in  the  surface. 

7.  Every  surface,  which  is  not  a  plane  surface,  or  composed 
of  plane  surfaces,  is  a  curved  surface. 

8.  A  solid  or  body  is  that  which  has  length,  breadth,  and 
thickness ;  and  therefore  combines  the  three  dimensions  of 
extension. 


10 


GEOMETRY 


9.  When  two  straight  lines,  AB,  AC,  meet 
each  other,  their  inclination  or  opening  is  call- 
ed an  angle^  which  is  greater  or  less  as  the 
lines  are  more  or  less  inchned  or  opened.  The 
point  of  intersection  A  is  the  vertex  of  the  ^, 
angle,  and  the  lines  AB,  AC,  are  its  sides. 

The  angle  is  sometimes  designated  simply  by  the  letter  at 
the  vertex  A  ;•  sometimes  by  the  three  letters  BAC,  or  CAB, 
the  letter  at  the  vertex  being  always  placed  in  the  middle. 

Angles,  like  all  other  quantities,  are  susceptible  of  addition, 
subtraction,  multipllGation,  and  division. 


Thus  the  angle  DCE  is  the  sum  of 
the  two  angles  DCB,  BCE ;  and  the  an- 
gle DCB  is  the  difference  of  the  two  £^ 
angles  DCE,  BCE. 


10.  When  a  straight  line  AB  meets  another 
straight.  line  CD,  so  as  to  make  the  adjacent 
angles  BAC,  BAD,  equal  to  each  other,  each 
of  these  angles  is  called  a  right  angle  ;  and  the 
line  AB  is  said  to  be  peiyendicular  to  CD.        ; 


A 


D 


11.  Every  angle  BAC,  less  than  a^> 
right  angle,  is  an  acute  angle ;  and 
every  angle  DEF,  greater  than  a  right 
angle,  is  an  obtuse  angle. 


-r 


12.  Two  lines  are  said  to  hQ  parallel,  when  — 
being  situated  in  the  same  plane,  they  cannot 
meet,  how  far  soever,  either  way,  both  of  them 
be  produced. 

13.  A  "plane  figure  is  a  plane  terminated  on 
all  sides  by  lines,  either  straight  or  curved. 

If  the  lines  are  straight,  the  space  they  enclose 
is  called  a  rectilineal  figure,  or  polygon,  and  the 
lines  themselves,  taken  together,  form  the  contour, 
or  perimeter  of  the  polygon. 

14.  The  polygon  of  three  sides,  the  simplest  of  all,  is  called 
a  triangle  ;  that  of  four  sides,  a  quadrilateral;  that  of  five,  a 
pentagon;  that  of  six,  a  hexagon  ;  that  of  seven,  a  heptagon; 
that  of  eight,  an  octagon  ;  that  of  nine,  a  nonagon;  that  of  ten,  a 
decagon  ;  and  that  of  twelve,  a  dodecagon. 


^M:: 


BOOK  I. 


11 


15.  An  equilateral  iriaugle  is  one 'which  has  its  three  sides 
equal  ;'an  isosceles  triangle,  one  which  h^s  two  of  its  sides 
equal ;  a  scalene  triangle,  one  which  has  its  three  sides  unequal 

16.  A  right-angled  triangle  is  one  which 
has  a  right  angle.  The  side  opposite  the 
right  angle  is  called  the  liypothenUse.  Thus, 
in  the  triangle  ABC,  right-angled  at  A,  the 
side  BC  is  the  hypothenuse. 

17.  Among  the  quadrilaterals,  we  distinguish  : 

The  square,  which  has  its  sides  equal,  and  its  an- 
gles right-angles. 


The  rectangle,  which  has  its  angles  right  an- 
gles,-without  having  its  sides  equal. 

The  parallelogram,  or  rhomboid,  which 
has  its  opposite  sides  parallel.  / 


The  rhombus,  or  lozenge,  which  has  its  sides  equal, 
without  having  its  angles  right  angles. 


And  lastly,  the  trapezoid,  only  two  of  whose  sides 
are  parallel. 


18.  A  diagonal  is  a  line  which  joins  the  ver- 
tices  of  two  angles  not  adjacent  to  each  other. 
Thus,  AF,  AE,  AD,  AC,  are  diagonals.  ^^ 


19.  An  equilateral  polygon  is  one  which  has  all  its  sides 
equal ;  an  equiangular  polygon,  one  which  has  all  its  angles 
equal. 

20.  Two  polygons  are  mutually  equiloieral,  when  they  have 
their  sides  equal  each  to  each,  an.d  placed  in  the  same  order ; 


'^^  Oft  xm    ■ 


12  GEOMETRY. 

that  is  to  say,  when  following  their  perimeters  in  the  same  di- 
rection, the  first  side  of  the  one  is  equal  to  the  first  side  of  the 
other,  the  second  of  the  one  to  the  second  of  the  other,  the 
third  to  the  third,  and  so  on.  The  phrase,  mutually  equian- 
gular, has  a  corresponding  signification,  with  respect  to  the 
angles. 

In  both  cases,  the  equal  sides,  or  the  equal  angles,  are  named 
homologous  sides  or  angles. 


Definitions  of  terms  employed  in  Geometry, 

An  axiom  is  a  self-evident  proposition. 

A  theorem  is  a  truth,  which  becomes  evident  by  means  of  a 
train  of  reasoning  called  a  demonstration. 

A  problem  is  a  question  proposed,  which  requires  a  solu- 
tion, 

A  lemma  is  a  subsidiary  truth,  employed  for  the  demonstra- 
tion of  a  theorem,  or  the  solution  of  a  problem. 

The  common  name,  proposition,  is  applied  indifferently,  to 
theorems,  problems,  and  lemmas. 

A  corollary  is  an  obvious  consequence,  deduced  from  one  or 
several  propositions. 

A  scholium  is  a  remark  on  one  or  several  preceding  propo-' 
sitions,  which  tends  to  point  out  their  connexion,  their  use,  their 
restriction,  or  their  extension. 

A  hypothesis  is  a  supposition,  made  either  in  the  enunciation 
of  a  proposition,  or  in  the  course  of  a  demonstration. 


Explanation  of  the  symbols  to  be  employed. 

The  sign  =  is  the  sign  of  equality;  thus,  the  expression 
A=B,  signifies  that  A  is  equal  to  B. 

To  signify  that  A  is  smaller  than  B,  the  expression  A<B 
is  used. 

To  signify  that  A  is  greater  than  B,  the  exp#i'ession,A>B 
is  used  ;  the  smaller  quantity  being  always  at  the  vertex  of  the 
angle. 

The  sign  +  is  called  plus :  it  indicates  addition. 

The  sign  —  is  called  jninus :  it  indicates  subtraction. 
Thus,  A  +  B,  represents  the  sum  of  the  quantities  A  and  B; 
A — B  represents  their  diiference,  or  what  remains  after  B  is 
taken  from  A  ;  and  A — B  +  C,  or  A  +  C — B,  signifies  that  A 
and  C  are  to  be  added  together,  and  that  B  is  to  be  subtracted 
from  their  «jum. 


BOOK  I.  13 

The  sign  x  indicates  multiplication  :  thus,  A  x  B  represents 
tlie  product  of  A  and  B.  Instead  of  the  sign  x ,  a  point  is 
Bometimes  employed  ;  thus,  A.B  is  the  same  thing  as  A  x  B. 
The  same  product  is  also  designated  without  any  intermediate 
«ign,  by  AB  ;  but  this  expression  should  not  be  employed,  when 
there  is  any  danger  of  confounding  it  with  that  of  the  line  AB, 
which  expresses  the  distance  between  the  points  A  and  B. 

The  expression  A  x  (B  +  C — D)  represents  the  product  of 
A  by  the  quantity  B  +  C — D.  If  A  +  B  were  to  be  multiplied 
by  A — B  +  C,  the  product  would  be  indicated  thus,  (A  +  B)x 
(A — B  +  C),  whatever  is  enclosed  within  the  curved  lines,  being 
considered  as  a  single  quantity. 

A  number  placed  before  a  hne,  or  a  quantity,  serves  as  a 
multiplier  to  that  line  or  quantity ;  thus,  3AB  signifies  that 
the  line  AB  is  taken  three  times ;  i  A  signifies  the  half  of  the 
angle  A. 

The  square  of  the  line  AB  is  designated  by  AB^ ;  its  cube 
by  AB^  What  is  meant  by  the  square  and  cube  of  a  line,  will 
be  explained  in  its  proper  place. 

The  sign  V  indicates  a  root  to  be  extracted  ;  thus  -^2 
means  the  square-root  of  2  ;  \/ A  x  B  means  the  square-root  of 
the  product  of  A  and  B. 

Axioms. 

1.  Things  which  are  equal  to  the  same  thing,  are  equal  to 
each  other. 

2.  If  equals  be  added  to  equals,  the  wholes  will  be  equal. 

3.  If  equals  be  taken  from  equals,  the  remainders  will  be 
equal. 

4.  If  equals  be  added  to  unequals,  the  wholes  will  be  un- 
equal. 

5.  If  equals  be  taken  from  unequals,  the  remainders  will  be 
unequal. 

6.  Things  which  are  double  of  the  same  thing,  are  equal  to 
each  other. 

7.  Things  which  are  halves  of  the  same  thing,  are  equal  to 
each  other. 

8.  The  whole  is  greater  than  any  of  its  parts. 

9.  The  whole  is  equal  to  the  sum  of  all  its  parts. 

10.  All  right  angles  are  equal  to  each  other. 

1 1  From  one  point  to  another  only  one  straight  line  can  be 
drawn. 

12.  Through  the  same  point,  only  one  straight  line  can  be 
drawn  Which  shall  be  parallel  to  a  given  line. 

13.  Magnitudes,  which  being  applied  to  each  other,  coincide 
throughout  their  whole  extent,  are  equal. 

B 


14 


GEOMETRY. 


PROPOSITION   I.    THEOREM. 

Ij  one  straight  line  meet  another  straight  line,  the  sum  of  the 
two  adjacent  angles  will  he  equal  to  two  right  angles. 

Let  the  straight  line  DC  meet  the  straight 
line  AB  at  C,  then  will  the  angle  ACD  + 
the  angle  DCB,  be  equal  to  two  right  angles. 

At  the  point  C,  erect  CE  perpendicular  to 

AB.     The  angle  ACD  is  the  sum  of  the  an-v p         ^ 

gles  ACE,  ECD:  therefore  ACD  +  DCB  is  ^         "^ 

the  sum  of  the  three  angles  ACE,  ECD,  DCB :  but  the  first 
of  these  three  angles  is  a  right  angle,  and  the  other  two 
make  up  the  right  angle  ECB ;  hence,  the  sum  of  the  two  an- 
gles ACD  and  DCB,  is  equal  to  two  right  angles. 

Cor.  1.     If  one  of  the  angles  ACD,  DCB,  is  a  right  angle, 
the  other  must  be  a  right  angle  also. 

Cor.  2.  If  the  line  DE  is  perpendicular 
to  AB,  reciprocally,  AB  will  be  perpendicu- 
lar to  DE. 

For,  since  DE  is  perpendicular  to  AB,  the 
angle  ACD  must  be  equal  to  its  adjacent  an- 
gle DCB,  and  both  of  them  must  be  right 
angles  (Def.  10.).  But  since  ACD  is  a 
right  angle,  its  adjacent  angle  ACE  must  also  be  a  right  angle 
(Cor.  1.).  Hence  the  angle  ACD  is  equal  to  the  angle  ACE, 
(Ax.  10.)  :  therefore  AB  is  perpendicular  to  DE. 

Cor.  3.  The  sum  of  all  the  successive 
angles,  BAC,  CAD,  DAE,  EAF,  formed 
on  the  same  side  of  the  straight  line  BF, 
is  equal  to  two  right  angles. ;  for  their  sum 
is  equal  to  that  of  the  two  adjacent  an- 
gles, BAC,  CAR 


D 


33 


PROPOSITION  II.    THEOREM. 

2\vo  straight  lines,  which  have  two  points  common,  coincide  with 
each  other  throughout  their  whole  extent,  and  form  one  ana 
the  same  straight  line. 

Let  A  and  B  be  the  two  common 
points.  In  the  first  place  it  is  evident 
that  the  two  lines  must  coincide  entirely 
between  A  and  B,  for  otherwise  there 
would  be  two  straight  lines  between  A  t"— ng" 
and  B,  which  is  impossible  (Ax.4 1 ) .  Sup- 


BOOK  I.  15 

pose,  however,  that  on  being  produced,  these'  lines  begin  to 
separate  at  C,  the  one  becoming  CD,  the  other  CE.  From 
the  point  C  draw  the  hne  CF,  making  with  AC  the  right  angle 
ACF.  Now,  since  ACD  is  a  straight  line,  the  angle  FCD  will 
be  a  right  angle  (Prop.  I.  Cor.  1.);  and  since  ACE  is  a  straight 
line,  the  angle  FCE  will  likewise  be  a  right  angle.  Hence,  the 
angle  FCD  is  equal  to  the  angel  FCE  (Ax.  10.);  which  can 
only  be  the  case  when  the  lines  CD  and  CE  coincide :  there- 
fore, the  straight  lines  which  have  two  points  A  and  B  com- 
mon, cannot  separate  at  any  point,  when  produced ;  hence  they 
form  one  and  the  same  straight  line. 

PROPOSITION  III.    THEOREM. 

If  a  straight  line  meet  two  other  straight  lines  at  a  common 
point,  making  the  sum  of  the  two  adjacent  angles  equal  to  two 
right  angles,  the  two  straight  lines  which  are  met,  will  form 
one  and  the  same  straight  line. 

Let  the  straight  line  CD  meet  the 
two  lines  AC,  CB,  at  their  common 
point  C,  making  the  sum  of  iht;  two 
adjacent  angles  DCA,  DCB,  equal  to  j^ 
two  right  angles ;  then  will  CB  be  the 
prolongation  of  AC,  or  AC  and  CB 
will  form  one  and  the  same  straight  line. 

For,  if  CB  is  not  the  prolongation  of  AC,  let  CE  be  that  pro- 
longation: then  the  line  ACE  being  straight,  the  sum  of  the 
angles  ACD,  DCE,  will  be  equal  to  two  right  angles  (Prop.  I.). 
But  by  hypothesis,  the  sum  of  the  angles  ACD,  DCB,  is  also 
equal  to  two  right  angles :  therefore,  ACD  +  DCE  nmst  be  equal 
to  ACD + DCB  ;  and  taking  away  the  angle  ACO  from  each, 
there  remains  the  angle  DCE  equal  to  the  angle  "DCB,  which 
can  only  be  the  case  when  the  lines  CE  and  C !>  coincide  ; 
hence,  AC,  CB,  form  one  and  the  same  straight  li.  ;. 

PROPOSITION  IV.    THEOREM. 

When  two  straight  lines  intersect  each  other,  the  ojrp   'I'e  or  Dcr- 
tical  angles,  which  they  foam,  are  equciL 


«♦ 


16 


GEOMETRY. 


Let  AB  and  DE  be  two  straight^ 
lines,  intersecting  each  other  at   C  ; 
then  will  the  angle  ECB  be  equal  to 
the  angle  ACD,  and  the  angle  ACE  to 
the  angle  DCB.  T)  B" 

For,  since  the  straight  line  DE  is  met  by  the  straight  line 
AC,  the  sum  of  the  angles  ACE,  ACD,  is  equal  to  two  right 
angles  (Prop.  L) ;  and  since  the  straight  line  AB,  is  met  by  the 
straight  line  EC,  the  sum  of  the  angles  ACE  and  ECB,  is  equal 
to  two  right  angles  :  hence  the  sum  ACE  +  ACD  is  equal  to 
the  sum  ACE  +  ECB  (Ax.  1.).  Take  away  from  both,  the  com- 
mon angle  ACE,  there  remains  the  angle  ACD,  equal  to  its 
opposite  or  vertical  angle  ECB  (Ax.  3.). 

Scholium.  The  four  angles  formed  about  a  point  by  two 
straight  lines,  which  intersect  each  other,  are  together  equal  to 
four  right  angles  :  for  the  sum  of  the  two  angles  ACE,  ECB, 
is  equal  to  two  right  angles  ;  and  the  sum  of  the  other  two, 
ACD,  DCB,  is  also  equal  to  two  right  angles  :  therefore,  the 
sum  of  the  four  is  equal  to  four  right  angles. 

In  general,  if  any  number  of  straight  lines 
CA,  CB,  CD,  &c.  meet  in  a  point  c,  the  >s, 
sum  of  all  the  successive  angles  ACB,BCD, 
DCE,  ECF,  FCA,  will  be  equal  to  four 
right  angles :  for,  if  four  right  angles  were 
formed  about  the  point  C,  by  two  lines  per- 
pendicular to  each  other,  the  same  space 
would  be  occupied  by  the  four  right  angles,  as  by  the  succes- 
sive angles  ACB,  BCD,  DCE,  ECF,  FCA. 


PROPOSITION  V.    TliEORlIM. 

If  two  triangles  have  two  sides  and  the  included  angle  of  the  one, 
equal  to  two  sides  and  the  included  angle  of  the  other,  each  to 
each,  the  two  triangles  will  be  equal 

Let  the  side  ED  be  equal 
to  the  side  BA,  the  side  DF 
to  the  side  AC,  and  the  an- 
gle D  to  the  angle  A ;  then 
will  the  triangle  EDF  be 
equal  to  the  triangle  BAC.  ^^ . — -=  — ^ 

For,  these  triangles  may  be  so  applied  to  each  other,  that  they 
shall  exactly  coincide.  Let  the  triangle  EDF,  be  placed  upon 
the  triangle  BAC,  so  that  the  point  E  shall  fall  upon  B,  and  the 
side  ED  on  the  equal  side  BA;  then,  since  the  angle  D  is  equal 
to  the  angle  A,  the  side  DF  will  take  the  direction  AC.     But 


% 


BOOK  I.  1^ 

DF  is  equal  to  AC  ;  therefore,  the  point  F  will  fall  on  C,  and 
the  third  side  EF,  will  coincide  with  the  third  side  BC  (Ax.  11.): 
therefore,  the  triangle  EDF  is  equal  to  the  triangle  BAG 
(Ax.  13.). 

Cor,  When  two  triangles  have  these  three  things  equals 
namely,  the  side  ED=BA,  the  side  DF=AC,  and  the  angle 
D=A,  the  remaining  three  are  also  respectively  equal,  namely, 
the  side  EF=BC,  the  angle  E=B,  and  the  angle  F=C 


PROPOSITION  VI.    THEOREM. 

If  two  triangles  have  two  angles  and  the  included  side  of  the  one, 
equal  to  two  angles  and  the  included  side  of  the  other,  each  to 
each,  the  two  triangles  will  be  equal. 

Let  the  angle  E  be  equal 
to  the  angle  B,  the  angle  F 
to  the  angle  C,  and  the  in- 
cluded side  EF  to  the  in- 
cluded side  BC ;  then  will 

the  triangle  EDF  be  equal 

to  the  triangle  BAC.  ^  ^^  C 

For  to  apply  the  one  to  the  other,  let  the  side  EF  be  placed 
on  its  equal  BC,  the  point  E  falHng  on  B,  and  the  point  F  on 
C ;  then,  since  the  angle  E  is  equal  to  the  angle  B,  the  side  ED 
will  take  the  direction  BA ;  and  hence  the  point  D  will  be  found 
somewhere  in  the  hne  BA.  In  like  manner,  since  the  angle 
F  is  equal  to  the  angle  C,  the  line  FD  will  take  the  direction 
CA,  and  the  point  D  will  be  found  somewhere  in  the  line  CA. 
Hence,  the  point  D,  falhng  at  the  same  time  in  the  two  straight 
lines  BA  and  CA,  must  fall  at  their  intersection  A:  hence,  the 
two  triangles  EDF,  BAC,  coincide  with  each  other,  and  are 
therefore  equal  (Ax.  13.). 

Cor.  Whenever,  in  two  triangles,  these  three  things  are  equal, 
namely,  the  angle  E=B,  the  angle  F=C,  and  the  included  side 
EF  equal  to  the  included  side  BC,  it  may  be  inferred  that  the 
remaining  three  are  also  respectively  equal,  namely,  the  angle 
D=A,  the  side  ED=BA,  and  the  side  DF=AC. 

Scholium,  Two  triangles  are  said  to  be  €qual,  when  being 
applied  to  each  other,  they  will  exactly  coincide  (Ax.  13.). 
Hence,  equal  triangles  have  their  like  parts  equal,  each  to  each, 
since  those  parts  must  coincide  with  each  other.  The  converse 
of  this  proposition  is  also  true,  namely,  that  two  triangles  which 
have  all  the  parts  of  the  one  equal  to  the  parts  of  the  other,  each 

B* 


18 


GEOMETRY. 


to  each,  are  equal ;  for  they  may  be  applied  to  each  other,  and 
the  equal  parts  will  mutually  coincide. 

PROPOSITION  VII.    THEOREM. 

The  sum  of  any  two  sides  of  a  triangle,  is  greater  than  the 
third  side. 

Let  ABC  be  a  triangle  :  then  will  the 
sum  of  two  of  its  sides,  as  AC,  CB,  be 
greater  than  the  third  side  AB. 

For  the  straight  line  AB  is  the  short- 
est distance  between  the  points  A  and 
B  (Def.  3.) ;  hence  AC  +  CB  is  greater 
than  AB, 


PROPOSITION  VIII.     THEOREM. 

If  from  any  point  within  a  triangle,  two  straight  lines  he  drawn 
to  the  extremities  of  either  side,  their  sum  will  he  less  than  the 
sum  of  the  two  other  sides  of  the  triangle. 

Let  any  point,  as  O,  be  taken  within  the  trian- 
gle BAC,  and  let  the  lines  OB,  OC,  be  drawn 
to  the  extremities  of  either  side,  as  BC  ;  then 
willOB  +  OC<BA+AC. 

Let  BO  be  produced  till  it  meets  the  side  AC 
in  D  :  then  the  line  OC  is  shorter  than  OD  +  DC^ 
(Prop.  VII.):  add  BO  to  each,  and  we  have  BO  +  OC<BO+ 
OD  +  DC  (Ax.  4.),  or  BO  +  OC<BD  +  DC. 

Again,  BD<BA+ AD:  add  DC  to  each,  and  we  have  BD  + 
DC<BA  +  AC.  But  it  has  just  been  found  that  BO  +  OC< 
BD  +  DC  ;  therefore,  still  more  is  BO  +  OC<BA+AC. 

* 

PROPOSITION    IX.    THEOREM. 

if  two  triangles  have  two  sides  of  the  one  equal  to  two  sides  of  the 
other,  each  to  each,  and  the  included  angles  unequal,  the  third 
sides*will  he  unequal ;  and  the  greater  side  will  belong  to  the 
triangle  which  has  the  greater  included  angle. 

Let  BAC  and  EDF 
bd  two  triangles,  having 
the  sideAB=DE,  AC 
=DF,  and  the  angle 
A>D;  then  will  BC> 
EF. 

Make  the  angle  C  AG^ 
=  D;   take    AG=:DE, 
and  draw  CG.      The 


BOOK  I. 


19 


triangle  GAC  is  equal  to  DEF,  since,  by  construction,  they 
have  an  equal  angle  in  each,  contained  by  equal  sides,  (Prop. 
V.) ;  therefore  CG  is  equal  to  EF.  Now,  there  may  be  three 
cases  in  the  proposition,  according  as  the  point  G  falls  without 
the  triangle  ABC,  or  upon  its  base  BC,  or  within  it. 

First  Case.  The  straight  line  GC<GI  +  IC,  and  the  straight 
line  AB<AI  +  IB;  therefore,  GC  +  AB<  GI  +  AI  +  IC  +  IB, 
or,  which  is  the  same  thing,  GC  +  AB<AG+BC.  Take  away 
AB  from  the  one  side,  and  its  equal  AG  from  the  other ;  and 
there  remains  GC<BC  (Ax.  5.)  ;  but  we  have  found  GC=EF, 
therefore,  BC>EF. 


Second  Case,  If  the  point  G 
fail  on  the  side  BC,  it  is  evident 
that  GC,  or  its  equal  EF,  will  be 
shorter  than  BC  (Ax.  8.). 


Third  Case,  Lastly,  if  the  point  G 
fall  within  the  triangle  BAC,  we  shall 
have,  by  the  preceding  theorem,  AGj^l* 
GC<AB  +  BC;  and,  taking  AG  from 
tlie  one,  and  its  equal  AB  from  the  other, 
ihere  will  remain  GC<BCorBC>EF.B 


Scholium.  Conversely,  if  two  sides 
BA,  AC,  of  the  triangle  BAC,  are  equal 
to  the  two  ED,  DF,of  the  triangle  EDF, 
each  to  each,  while  the  third  side  BC  of 
the  first  triangle  is  greater  than  the  third 
side  EF  of  the  second  ;  then  will  the  an- 
gle BAC  of  the  first  triangle,  be  greater 
than  the  angle  EDt^  of  the  second. 

For,  if  not,  the  angle  BAC  must  be  equal  to  EDF,  or  less 
than  it.  In  the  first  case,  the  side  BC  would  be  equal  to  EF, 
(Prop.  V.  Cor.) ;  in  the  second,  CB  would  be  less  than  EF ;  but 
either  of  these  results  contradicts  the  hypothesis ;  therefore,  BAC 
is  greater  than  EDF. 


PROPOSITION  X.    THEOREM. 


If  two  triangles  have  the  three  sides  of  the  one  equal  to  the  three 
sides  of  the  other,  each  to  each,  the  three  angles  will  also  he 
equal,  each  to  each,  and  the  triangles  themselves  will  be  equal. 


m- 


20  GEOMETRY 


Let  the  side  ED=BA, 
ihe  side  EF=BC,  and  the 
side  DF=AC  ;  then  will 
the  angle  D=A,  the  angle 

E=B,  and  the   angle   F 

=  C.  E  TB  C 

For,  if  the  angle  D  were  greater  than  A,  while  the  sides 
ED,  DF,  were  equal  to  BA,  AC,  each  to  each,  it  would  fol- 
low, by  the  last  proposition,  that  the  side  EF  must  be  greater 
than  BC  ;  and  if  the  angle  D  were  less  than  A,  it  would  follow, 
that  the  side  EF  must  be  less  than  BC  :  but  EF  is  equal  to  BC, 
by  hypothesis  ;  therefore,  the  angle  D  can  neither  be  greater 
nor  less  than  A  ;  therefore  it  must  be  equal  to  it.  In  the  same 
manner  it  may  be  shown  that  the  angle  E  is  equal  to  B,  and 
the  angle  F  to  C :  hence  the  two  triangles  are  equal  (Prop. 
VI.  Sch.). 

Scholium.  It  may  be  observed  that  the  e(|Ual  angles  lie  op- 
posite the  equal  sides :  thus,  the  equal  angles  D  and  A,  lie  op- 
posite the  equal  sides  EF  and  BC. 

PROPOSITION  XI.    THEOREM. 

In  an  isosceles  triangle,  the  angles  opposite  the  equal  sides 
are  equal. 

Let  the  side  BA  be  equal  to  the  side  AC ;  then 
will  the  angle  C  be  equal  to  the  angle  B. 

For,  joife  the  vertex  A,  and  D  the  middle  point 
of  the  base  BC.  Then,  the  triangles  BAD,  DAC, 
will  have  all  the  sides  of  the  one  equal  to  those 
of  the  other,  each  to  each  ;  for  BA  is  equal  to  AC,]^ 
by  hypothesis ;  AD  is  common,  and  BD  is  equal 
to  DC  by  construction :  therefore,  by  the  last  proposition,  the 
angle  B  is  equal  to  the  angle  C. 

Cor.  An  equilateral  triangle  is  likewise  equiangular,  that  is 
to  say,  has  all  its  angles  equal. 

Scholium.  The  equality  of  the  triangles  BAD,  DAC,  proves 
also  that  the  angle  BAD,  is  equal  to  DAC,  and  BDA  to  ADC, 
hence  the  latter  two  are  right  angles ;  therefore,  the  line  drawn 
from  the  vertex  of  an  isosceles  triangle  to  the  middle  point  of  its 
base,  is  perpendicular  to  the  base,  and  divides  tlie  angle  at  the 
vertex  into  two  equal  parts. 

In  a  triangle  which  is  not  isosceles,  any  side  may  be  assumed 
indifferently  as  the  base ;  and  the  vertex  is,  in  that  case,  the 
vertex  of  the  opposite  angle.  In  an  isosceles  triangle,  however 


BOOK  I.  21 

(■• 
that  side  is  generally  assumed  as  the  base,  which  is  not  equnl 
to  either  of  the  other  two. 


PROPOSITION  XII.    THEOREM. 

Conversely,  if  two  angles  of  a  triangle  are  equal,  the  sides  oppo- 
site them  are  also  equal,  and  the  triangle  is  isosceles. 

Let  the  angle  ABC  be  equal  to  the  angle  ACB ; 
then  will  the  side  AC  be  equal  to  the  side  AB. 

For,  if  these  sides  are  not  equal,  suppose  AB 
to  be  the  greater.  Then,  take  BD  equal  to  AC, 
and  draw  CD.  Now,  in  the  two  triangles  BDC, 
BAC,  we  have  BD=:AC,  by  construction;  the 
angle  B  equal  to  the^ngle  ACB,  by  hypothesis  ;b^ 
and  the  /side  BC  common  :  therefore,  the  two 
txidriglefe,  BDC,  BAC,  have  two  sides  and  the  included  angle  in 
the  on^,  equal  to  two  sides  and  the  included  angle  in  the  other, 
each  to  each  :  hence  they  are  equal  (Prop.  V.).  i  But  the  part 
«annot  be  equal  to  the  whole  (Ax.  8.) ;  hencej'^rliej'e  is^no 
'nequahty  ijutw^en  thq  sides  BA,  AC  ;  therefore,  the  triangle 
C  is  isosceles. 


..^ 


PROPOSITION  XIII.    THEOREM. 


■g^y^The  greater  side  of  every  triangle  is  opposite  to  the  greater  an- 
'     gle ;  and  conversely,  the  greater  angle  is  opposite  to  the 
greater  side. 

First,  Let  the  angle  C  be  greater  than  the  angle 
B  ;  then  will  the  side  AB,  opposite  C,  be  greater 
than  AC,  opposite  B. 

For,  make  the  angle  BCD=B.     Then,  in  the 
triangle  CDB,  we  shall  haveCD-BD  (Prop.XIL). 
Now,  the  side  AC < AD  +  CD;  butAD+CD=C' 
AD  +  DB=AB :  therefore  AC<  AB. 

Secondly,  Suppose  the  side  AB>AC;  then  will  the  angle  C, 
opposite  to  AB,  be  greater  than  the  angle  B,  opposite  to  AC. 

For,  if  the  angle  C<B,  it  follows,  from  what  has  just  been 
proved,  that  AB<AC;  which  is  contrary  to  the  hypothesis.  It 
the  angle  C=B,  then  the  side  AB=AC  (Prop.  XJI.);  which  is 
also  contrary  to  the  supposition.  Therefore,  when  AB>AC, 
the  angle  C  must  be  greater  than  B. 


22  GEOMETRY. 


PROPOSITION  XIV.    THEOREM. 

From  a  given  point,  without  a  straight  line,  only  one  perpendicu- 
lar can  he  drawn  to  that  line. 

Let  A  be  the  point,  and  DE  the  given 
line. 

Let  us  suppose  that  we  can  draw  two 
perpendiculars,  AB,  AC.  Produce  either 
of  them,  as  AB,  till  BF  is  equal  to  AB,  and  D- 
draw  FC.  Then,  the  two  triangles  CAB, 
CBF,  will  be  equal:  for,  the  angles  CBA, 
and  CBF  are  right  angles,  the  side  CB  is  Np 

common,  and  the  side  AB  equal  to  BF,  by  construction  ;  there- 
fore, the  triangles  are  equal,  and  the  angle  ACB=:BCF  (Prop. 
V.  Cor.).  But  the  angle  ACB  is  a  right  angle,  by  hypothesis  ; 
therefore,  BCF  must  likewise  be  a  right  angle.  But  if  the  adja- 
cent angles  BCA,  BCF,  are  together  equal  to  two  right  angles, 
ACF  must  be*  a  straight  line  (Prop.  IIL) :  from  whence  it  fol- 
lows, that^tween  the  same  two  points,  A  and  F,  two  straight 
lines  can  be  drawn,  which  is  impossible  (Ak.  1 1,).-  iience,  two 
perpendiculars  cannot  be  drawn  from  the  same  point  to  the^^. 
same  straight  line. 

Scholium,    At  a  given  point  C,  in  the  line  j;  i 

AB,  it  is  equally  impossible  to  erect  two  per- 
pendiculars to  that  line.  For,  if  CD,  CE, 
were  those  two  perpendiculars,  the  angles 
BCD,  BCE,  would  both  be  right  angles:— 
hence  they  would  be  equal  (Ax.  10.);  and 
the  line  CD  would  coincide  withCE;  otherwise,  a  part  would 
be  equal  to  the  whole,  which  is  impossible  (Ax.  8.). 


PROPOSITION  XV.    THEOREM. 

If  from  a  point  without  a  straight  line,  a  perpendicular  he  let 
fall  on  the  line,  and  ohlique  lines  he  drawn  to  different  points : 

1st,  The  perpendicular  will  he  shorter  than  any  ohlique  line, 

2d,  Any  two  ohlique  lines,  drawn  on  different  sides  of  the  perpen- 
dicular, cutting  off  equal  distances  on  the  other  line,  will  he 
equal, 

Sd,  Of  two  ohlique  lines,  drawn  at  pleasure,  that  which  is  farther 
from  the  perpendicular  will  he  the  longer. 


BOOK  I.  23 

Let  A  be  the  given  point,  DE  the  given 
line,  AB  the  perpendicular,  and  AD,  AC, 
AE,  the  oblique  lines. 

Produce  the  perpendicular  AB  till  BF 
is  equal  to  AB,  and  draw  FC,  FD.  D^ 

First.  The  triangle  BCF,  is  equal  to  the 
triangle  BCA,  for  tliey  have  the  right  angle 
CBF=CBA,  the  side  CB  common,  and  the  ^F 

side  BF=BA  ;  hence  the  third  sides,  CF  and  CA  are  equal 
(Prop.  V.  Cor.).  But  ABF,  being  a  straight  line,  is  shorter  than 
ACF,  which  is  a  broken  line  (Def.  3.) ;  therefore,  AB,  the  half 
of  ABF,  is  shorter  than  AC,  the  half  of  ACF ;  hence,  the  per- 
pendicular is  shorter  than  any  oblique  line. 

Secondly.  Let  us  suppose  BC=BE;  then  will  the  triangle 
CAB  be  equal  to  the  the  triangle  BAE  ;  for  BC=BE,the  side 
AB  is  common,  and  the  angle  CBA=ABE  ;  hence  the  sides 
AC  and  AE  are  equal  (Prop.  V.  Cor.)  :  therefore,  two  oblique, 
lines,  equally  distant  from  the  perpendicular,  are  equal. 

Thirdly.  In  the  triangle  DFA,  the  sum  of  the  lines  AC,  CF, 
is  less  than  the  sum  of  the  sides  AD,  DF  (Prop.  VIIL) ;  there- 
fore, AC,  the  half  of  the  line  ACF,  is  shorter  than  AD,  the  half 
of  the  line  ADF ;  therefore,  the  oblique  line,  which  is  farther 
from  the  perpendicular,  is  longer  than  the  one  which  is  nearer. 

Cor.  L  Th-e  perpendicular  measures  the  shortest  distance 
of  a  point  from  a  line. 

Cor.  2.  From  the  same  point  to  the  same  straight  line,  only 
two  equal  straight  lines  can  be  drawn ;  for,  if  there  could  be 
more,  we  should  have  at  least  two  equal  oblique  lines  on  the 
same  side  of  the  perpendicular,  which  is  impossible.  « 


PROPOSITION  XVI.    THEOREM. 

F  fj'om  the  middle  point  of  a  straight  line,  a  .perpendicular  he 

drawn  to  this  line  ; 
ist,  Every  point  of  th&  perpendicular  will  he  equally  distant 

from  the  extremities  of  the  line. 
2dy  Every  point,  without  the  perpendicular,  will  he  unequally  dis' 

tant  from  those  extremities. 


% 


%. 


24 


GEOMETRY. 


Let  AB  be  the  given  straight  line,  C  the 
middle  point,  and  ECF  the  perpendicular. 

First,  Since  AC=CB,  the  two  oblique  lines 
AD,  DB,  are  equally  distant  from  the  perpen- 
dicular, and  therefore  equal  (Prop.  XV.).  So, 
(ike wise,  are  the  two  oblique  hues  AE,  EB,  the 7^^ 
two  AF,  FB,  and  so  on.  Therefore  every  point 
m  the  perpendicular  is  equally  distant  from  the 
extremities  A  and  B. 

Secondly,  Let  I  be  a  point  out  of  the  perpen- 
dicular. If  lA  and  IB  be  drawn,  one  of  these  lines  will  cut 
the  perpendicular  in  D;  from  w^hich,  drawing  DB,  we  shall 
have  DB=DA.  But  the  straight  line  IB  is  less  than  ID+DB, 
and  ID  +  DB^ID  +  DA-IA;  therefore,  IB<IA;  therefore, 
every  point  out  of  the  perpendicular,  is  unequally  distant  from 
the  extremities  A  and  B. 

Cor.  If  a  straight  line  have  two  points  D  and  F,  equally  dis- 
tant from  the  extremities  A  and  B,  it  will  be  perpendicular  to 
AB  at  the  middle  point  C. 


PROPOSITION  XVII.    THEOREM. 

If  two  right  angled  triangles  have  the  hypothenuse  and  a  side  oj 
the  one,  equal  to  the  hypothenuse  and  a  side  of  the  other,  each  to 
each,  the  remaining  parts  will  also  he  equal,  each  to  each,  and 
the  triangles  themselves  will  ^^  equal. 

In  the  two  right  angled  * 
triangles  BAG,  EDF,  let  the 
hypothenuse  AG  =  DF,  and 
the  sideBA=ED:  then  will 
the  side  BC=EF,  the  angle  ^ 
A=D,  and  the  angle  C=F. 

If  the  side  BC  is  equal  to  EF,  the  like  angles  of  the  two 
triangles  are  equal  (Prop.  X.).  Now,  if  it  be  possible,  suppose 
these  two  sides  to  be  unequal,  and  that  BC  is  the  greater. 

On  BC  take  BG=:EF,  and  draw  AG.  Then,  in  the  two 
triangles  BAG,  DEF,  the  angles  B  and  E  are  equal,  being  right 
angles,  the  side  BA=ED  by  hypothesis,  and  the  side  BG=EF 
by  construction  :  consequently,  AG =DF  (Prop.  V.  Cor.).  But, 
by  hypothesis  AC=DF;  and  therefore,  AC=AG  (Ax.  1.). 
But  the  oblique  line  AC  cannot  be  equal  to  AG,  which  lies 
nearer  the  perpendicular  AB  (Prop.  XV.) ;  therefore,  BC  and 
EF  caDnot  be  unequal,  and  hence  the  angle  A=D,  and  the 
angle  C=F;  and  I'^ereforc,  the  triangles  are  equal  (Prop.  VI. 
Sch.). 


BOOK  I.        •  25 


PROPOSITION  XVIIl.    THEOREM. 

If  two  straight  lines  are  perpendicular  to  a  third  line,  they  will 
be  parallel  to  each  other :  in  other  words,  they  will  never  meet, 
how  far  soever  either  way,  both  of  them  be  produced. 

Let  the  two  lines  AC,  BD,  A.  C 

be  perpendicular  to  AB  ;  then 
will  they  be  parallel. 

For,  if  they  could  meet  in 
a  point  O,  on  either  side  of 
AB,  there  would  be  two  per- 


::::r:^o 


D 


pendiculars  OA,  OB,  let  fall  from  the  same  point  on  the  same 
straight  line;  which  is  impossible  (Prop.  XIV.). 


PROPOSITION  XIX.    THEOREM. 

If  two  straight  lines  meet  a  third  line,  maldng  the  sum  of  the 
interior  angles  on  the  same  side  of  the  line  me,i,  equal  to  two 
right  angles,  the  two  lines  will  be  parallel. 

Let  the  two  lines  EC,  BD,  meet 

the,  third  line  BA,  making  the  an-  ^      ^ 

gles  BAC,  ABD,  together  equal  to 
two  right  angles:  then  the  hnes 
EC,  BD,  will  be  parallel. 

From  G,  the  middle  point  'of 
BA,  draw  the  straight  line  EGF, 
perpendicular  to  EC.  It  will  also 
be  perpendicular  to  BD.  For,  the  sum  BAC  +  ABD  is  equal 
to  two  right  angles,  by  hypothesis ;  the  sum  BAC  +  BAE  is 
likewise  equal  to  two  right  angles  (Prop.  I.)  ;  and  taking  away 
BAC  from  both,  there  will  remain  the  angle  ABD=BAE. 

Again,  the  angles  EGA,  BGF,  are  equal  (Prop.  IV.) ;  there- 
fore, the  triangles  EGA  and  BGF,  have  each  a  side  and  two 
adjacent  angles  equal ;  therefore,  they  are  themselves  equal, 
and  the  angle  GEA  is  equal  to  the  angle  GFB  (Prop.  VI.  Cor.) : 
but  GEA  is  a  right  angle  by  construction  ;  therefore,  GFB  is  a 
right  angle ;  hence  the  two  Hnes  EC,  BD,  are  perpendicular  to 
the  same  straight  line,  and  are  therefore  parallel  (Prop.  XVIIL). 


26 


GEOMETRY. 


Scholium.  When  two  parallel 
straight  lines  AB,  CD,  are  met  by  a 
third  line  FE,  the  angles  which  are 
formed  take  particular  names.    . 

Interior  angles  on  the  same  side,  are 
those  which  lie  within  the  parallels,  ^- 
and  on  the  same  side  of  the  secant 
Mine :  thus,  0GB,  GOD,  are  interior 
angles  on  the  same  side  ;  and  so  also 
are  the  the  angles  OGA,  GOC. 

Alternate  angles  lie  within  the  parallels,  and  on  different 
sides  of  the  secant  line :  AGO,  DOG,  are  alternate  angles ; 
and  so  also  are  the  angles  COG,  BGO. 

Alternate  exterior  anglts  lie  without  the  parallels,  and  on  dif- 
ferent sides  of  the  secant  line :  EGB,  COF,  are  alternate  exte- 
rior angles ;  so  also,  are  the  angles  AGE,  FOD. 

Opposite  exterior  and  interior  angles  lie  on  the  same  side  of  the 
secant  line,  the  one  without  and  the  other  within  the  parallels, 
but  not  adjacent :  thus,  EGB,  GOD,  are  opposite  exterior  and 
interior  angles ;  and  so  also,  are  the  angles  AGE,  GOC. 

Cor.  1.  If  a  straight  line  EF,  meet  two  straight  lines  CD, 
AB,  making  the  alternate  angles  AGO,  GOD,  equal  to  each 
other,  the  two  lines  will  be  parallel.  For,  to  each  add  the  an- 
gle 0GB;  we  shall  then  have,  AGO  +  0GB = GOD  +  0GB  ; 
but  AGO  +  0GB  is  equal  to  two  right  angles  (Prop.  I.) ;  hence 
GOD  -4-  0GB  is  equal  to  two  right  angles :  therefore,  CD,  AB, 
are  parallel. 

Cor.  2.  If  a  straight  line  EF,  meet  two  straight  lines  CD, 
AB,  making  the  exterior  angle  EGB  equal  to  the  interior  and 
opposite  angle  GOD,the  two  lines  will  be  parallel.  For,  to  each 
add  the  angle  0GB:  we. shall  then  have  EGB  +  0GB = GOD 
+  0GB  :  but  EGB  +  0GB  is  equal  to  two  right  angles  ;  hence, 
GOD  4-  0GB  is  equal  to  two  right  angles ;  therefore,  CD,  AB, 
are  parallel. 

PROPOSITION  XX.    THEOREM. 

If  a  straight  line  meet  two  parallel  straight  lines,  the  sum  of  the 

interior  angles  on  the  same  side  will  he  equal  to  two  right  angles. 

Let  the  parallels  AB,CD,be 
met  by  tlie  secant  line  FE :  then 
will  OGB  +  GOD,  or  OGA  + 
GOC,  be  equal  to  two  right  an- 

For,  if  OGB  +  GOD  be  not 
equal  to  two  right  angles,  let 
IGH  be  drawn,  making  the  sum 
OGH+GOD  equal    to   two 


BOOK  1.  .  27 

right  angles ;  then  IH  and  CD  will  be  parallel  (Prop.  XIX.), 
and  hence  we  shall  have  two  lines  GB,  GH,  drawn  through 
the  same  point  G  and  parallel  to  CD,  which  is  impossible  (Ax. 
12.):  hence,  GB  and  GH  should  coincide,  and  0GB  +  GOD  is 
equal  to  two  right  angles.  In  the  same  manner  it  may  be  proved 
that  OGA+GOC  is  equal  to  two  right  angles. 

Cor.  1.  If  0GB  is  a  right  angle,  GOD  will  be  a  right  angle 
also:  therefore,  every  straight  line  perpendicular  to  one  of  two 
parallels,  is  perpendicular  to  the  other. 

Cor.  2.    If  a  straight  line  meet  two  ^ 

parallel  lines,  the  alternate  angles  will 
be  equal. 

Let  AB,  CD,  be  the  parallels,  and 
FE  the  secant  line.  The  sum  0GB + 
GOD  is  equal  to  two  right  angles.  But  ^ 
the  sum  OGB  +  OGA  is  also  equal  to 
two  right  angles  (Prop.  I.).  Taking 
from  each,  the  angle  OGB,  and  there 
remains  OGA=:GOD.  In  the  same  manner  we  may  prove  that 
GOC=OGB. 

Cor.  3.  If  a  straight  line  meet  two  parallel  lines,  the  oppo- 
site exterior  and  interior  angles  will  be  equal.  For,  the  sum 
OGB  +  GOD  is  equal  to  two  right  angles.  But  the  sum  OGB 
+  EGB  is  also  equal  to  two  right  angles.  Taking  from  each  the 
angle  OGB,  and  there  remains  GOD=EGB.  In  the  same 
manner  we  may  prove  that  AGE =GOC. 

Cor.  4,  We  see  that  of  the  eight  angles  formed  by  a  line 
cutting  two  parallel  lines  obliquely,  the  four  acute  angles  are 
equal  to  each  other,  and  so  also  are  the  four  obtuse  angles. 

PROPOSITION  XXI.    THEOREM. 

If  a  straight  line  meet  two  other  straight  lines,  making  the  sum  of 
the  interior  angles  on  the  same  side  less  than  two  rightangles^ 
the  two  lines  will  meet  if  sufficiently  produced* 

Letthe  line  EFmeet  the  two 
4ines  CD,  lil,  making  the  sum 
of  the  interior  angles  OGH, 
GOD,  less  than  two  right  an- 
gles :  then  will  IH  and  CD 
meet  if  sufficiently  produced. 

For,  if  they  do  not  meet  they 
are  parallel  (Def.  12.).  But  they 
are  not  parallel,  for  if  they  were, 
the  sum  of  the  interior  angles  OGH,  GOD,  w^ould  be  equal  to 
two  right  angles  (Prop.  XX.),  whereas  it  is  less  by  hypothesis : 
hence,  the  lines  IH,  CD,  are  not  parallel,  and  will  therefore 
meet  if  sufficiently  produced. 


28 


GEOMETRY. 


Co7\  It  is  evident  that  the  two  lines  IH,  CD,  will  meet  on 
that  side  of  EF  on  which  the  sum  f»f  the  two  angles  OGH, 
GOD,  is  less  than  two  right  angles 


PROPOSITION  XXII.    THEOREM. 

Two  straight  lines  which  are  parallel  to  a  third  line,  are  parallel 
to  each  other. 


.A. 


H 


■JS 


Let  CD  and  AB  be  parallel  to  the  third  line  EF ;  then  arc 
they  parallel  to  each  other. 

Draw  PQR  perpendicular  to  EF,  and 
cutting  AB,  CD.     Since  AB  is  parallel  to__ 
EF,  PR  will  be  perpendicular  to  AB  (Prop.E" 

XX.  Cor.  1.) ;   and  since  CD  is  parallel  to 

EF,  PR  will  for  a  like  reason  be  perpen-C 

dicular  to  CD.     Hence  AB  and  CD  are 

perpendicular  to   the  same  straight  line 
hence  they  are  parallel  (Prop.  XVIII.). 


t 


PROPOSITION  XXIII.    THEOREM. 
Two  parallels  are  every  where  equally  distant. 

Two  parallels  AB,  CD,  being  c  K 
given,  if  through  two  points  E 
and  F,  assumed  at  pleasure,  the 
straight  lines  EG,  FH,  be  drawn 
perpendicular  to  AB,these  straight^ 
lines  will  at  the  same  time  be 
perpendicular  to  CD  (Prop.  XX.  Cor.  1.)  :  and  we  are  now  to 
show  that  they  will  be  equal  to  each  othcv. 

If  GF  be  drawn,  the  angles  GFE,  FGII,  considered  in  refer- 
ence to  the  parallels  AB,  CD,  will  be  alternate  angles,  and 
therefore  equal  to  each  other  (Prop.  XX.  Cor.  2.).  Also,  the 
straight  lines  EG,  FH,  being  perpendicular  to  the  same  straight 
line  AB,  are  parallel  (Prop.  XVIII.) ;  and  the  angles  EGF, 
GFH,  considered  in  reference  to  the  parallels  EG,  FH,  will  be 
alternate  angles,  and  therefore  equal.  Hence  the  two  trian- 
gles EFG,  FGH,  have  a  common  side,  and  two  adjacent  angles 
in  each  equal ;  hence  these  triangles  are  equal  (Prop.  VI.)  ; 
therefore,  the  side  EG,  which  measures  the  distance  of  the 
parallels  AB  and  CD  at  the  point  E,  is  equal  to  the  side  FH. 
which  measures  the  distance  of  the  same  parallels  at  the 
point  F. 


BOOK  I.  29 


PROPOSITION  XXIV.    THEOREM. 


If  two  angles  have  their  sides  parallel  and  lying  in  the  same  di- 
rection,  the  two  angles  will  he  equal. 

Let  BAG  and  DEF  be  the  two  angles, 
having  AB  parallel  to  ED,  and  AC  to  EF ; 
then  will  the  angles  be  equal. 

For,  produce  DE,  if  necessary,  till  it 
meets  AC  in  G.     Then,  since  EF  is  par- 
allel to  GG,  the  angle  DEF  is  equal  to  ^ 
DGC   (Prop.  XX.  Cor.  3.);    and   since 
DG  is  parallel  to  AB,  the  angle  DGC  is  equal  to  BAG  ;  hence, 
the  angle  DEF  is  equal  to  BAG  (Ax.  1.). 

Scholium,  The  restriction  of  this  proposition  to  the  case 
where  the  side  EF  lies  in  the  same  direction  with  AG,  and  ED 
in  the  same  direction  with  AB,  is  necessary,  because  if  FE 
were  produced  towards  H,  the  angle  DEH  would  have  its  sides 
parallel  to  those  of  the  angle  BAG,  but  would  not  be  equal  to 
it.  In  that  case,  DEH  and  BAG  would  be  together  equal  to 
two  right  angles.  For,  DEH  +  DEF  is  equal  to  two  right  angles 
(Prop.  I.) :  but  DEF  is  equal  to  BAG :  hence,  DEH  +  BAG  is 
equal  to  two  right  angles. 

PROPOSITION  XXV.    THEOREM. 

In  every  triangle  the  sum  of  the  three  angles  is  equal  to  two 
■    right  angles. 

Let  ABG  be  any  triangle :  then  will  the  an- 
gle G+A+B  be  equal  to  two  right  angles. 

For,  produce  the  side  GA  towards  D,  and  at 
the  point  A,  draw  AE  parallel  to  BC.  Then, 
since  AE,  GB,  are  parallel,  and  GAD  cuts  them, 
the  exterior  angle  DAE  will  be  equal  to  its  inte-  C  AD 
rior  opposite  one  AGB  (Prop.  XX.  Cor.  3.) ;  in  like  manner, 
since  AE,  GB,  are  parallel,  and  AB  cuts  them,  the  alternate 
angles  ABG,  BAE,  will  be  equal :  hence  the  three  angles  of 
the  triangle  ABG  make  up  the  same  sum  as  the  three  angles 
GAB,  BAE,  EAD  ;  hence,  the  sum  of  the  three  angles  is  equal 
to  two  right  angles  (Prop.  L). 

Cor,  1.  Two  angles  of  a  triangle  being  given,  or  merely 
their  sum,  the  third  will  be  found  by  subtracting  that  sum  from 
two  right  angles. 


30  GEOMETRY. 

Cor.  2.  If  two  angles  of  one  triangle  a  re  respectively  equal 
to  two  angles  of  another,  the  third  angles  will  also  be  equal 
and  the  two  triangles  will  be  mutually  equiangular. 

Cor,  3.  In  any  triangle  there  can  be  but  one  right  angle  : 
for  if  there  were  two,  the  third  angle  must  be  nothing.  Still 
less,  can  a  triangle  have  more  than  one  obtuse  angle. 

Cor.  4.  In  every  right  angled  triangle,  the  sum  of  the  two 
acute  angles  is  equal  to  one  right  angle. 

Cor.  5.  Since  every  equilateral  triangle  is  also  equiangular 
(Prop.  XI.  Cor.),  each  of  its  angles  will  be  equal  to  the  third 
part  of  two  right  angles ;  so  that,  if  the  right  angle  is  expressed 
by  unity,  the  angle  of  an  equilateral  triangle  will  be  expressed 
byf.  ,         , 


Cor.  6.  In  every  triangle  ABC,  the  exterior  angle  BAD  is 
equal  to  the  sum  of  the  two  interior  opposite  angles  B  and  C. 
For,  AE  being  parallel  to  BC,  the  part  BAE  is  equal  to  the 
angle  B,  and  the  other  part  DAE  is  equal  to  the  angle  C. 


PROPOSITION  XXVI.    THEOREM. 

TJie  sum  of  all  the  interior  angles  of  a  polygon,  is  equal  to  two 
right  angles,  taken  as  many  times  less  two,  as  the  figure  has 
sides. 

Let  ABCDEFG  be  the  proposed  polygon. 
If  from  the  vertex  of  anyone  angle  A,  diagonals  jj^ 
AC,  AD,  AE,  AF,  be  drawn  to  the  vertices  of 
all  the  opposite  angles,  it  is  plain  that  the  poly-^^ 
gon  will  be  divided  into  five  triangles,  if  it  has 
seven  sides  ;  into  six  triangles,  if  it  has  eight;  and,  A. 
in  general,  into  as  many  triangles,  less  two,  as 
the  polygon  has  sides ;  for,  these  triangles  may  be  considered 
as  having  the  point  A  for  a  common  vertex,  and  for  bases,  the 
several  sides  of  the  polygon,  excepting  the  two  sides  whiqh  form 
the  angle  A.  It  is  evident,  also,  that  the  sum  of  all  the  angles 
in  these  triangles  does  not  differ  from  the  sum  of  all  the  angles 
in  the  polygon  :  hence  the  sum  of  all  the  angles  of  the  polygon 
is  equal  to  two  right  angles,  taken  as  many  times  as  there  are 
triangles  in  the  figure  ;  in  other  words,  as  there  are  units  in  the 
number  of  sides  diminished  by  two. 

Cor.  1.     The  sum  of  the  angles  in  a  quadrilateral  is  equal 
to  two  right  angles  multiplied  by  4 — 2,  which  amounts  to  four 


BOOK  I.  31 

right  angles :  hence,  if  all  the  angles  of  a  quadrilateral  are 
equal,  each  of  them  will  be  a  right  angle  ;  a  concluslop  which 
sanctions  the  seventeenth  Definition,  where  the  four  angles  of 
a  quadrilateral  are  asserted  to  be  right  angles,  in  the  case  of  the 
rectangle  and  the  square. 

Cor.  2.  The  sum  of  the  angles  of  a  pentagon  is  equal  to 
two  right  angles  multiplied  by  5 — 2,  which  amounts  to  six  right 
angles :  hence,  when  a  pentagon  is  equiangular,  .each  angle 
is  equal  to  the  fifth  part  of  six  right  angles,  or  to  |  of  one  right 
angle. 

Cor.  3.  The  sum  of  the  angles  of  a  hexagon  is  equal  to 
2  X  (6 — 2,)  or  eight  right  angles ;  hence  in  the  equiangular 
hexagon,  each  angle  is  the  sixth  part  of  eight  right  angles,  or  | 
of  one. 

Scholium.  When  this  proposition  is  applied  to 
polygons  which  have  re-emtrant  angles,  each  re- 
entrant angle  must  be  regarded  as  greater  than 
two  right  angles.  But  to  avoid  all  ambiguity,  we 
shall  henceforth  limit  our  reasoning  to  polygons 
with  salient  angles,  which  might  otherwise  be  named  convex 
polygons.  Every  convex  polygon  is  such  that  a  straight  line, 
drawn  at  pleasure,  cannot  meet  the  contour  of  the  polygon  in 
more  than  two  points. 


PROPOSITION  XXVII.    THEOREM. 

If  the  sides  of  any  polygon  he  produced  out,  in  the  same  direc- 
tion,  the  sum  of  the  exterior  angles  will  he  equal  to  four  right 
angles. 

Let  the  sides  of  the  polygon  ABCD- 
FG,  be  produced,  in  the  same  direction ; 
then  will  the  sum  of  the  exterior  angles 
a  +  &  +  c  -i-  d-^f-\-g,  be  equal  to  four  right 
angles. 

For,  each  interior  angle,  plus  its  ex- 
terior angle,  as  A  +  a,  is  equal  to  two 
right  angles  (Prop.  I.).  But  there  are 
as  many  exterior  as  interior  angles,  and  as  many  of  each  as 
there  are  sides  of  the  polygon  :  hence,  the  sum  of  all  the  inte- 
rior and  exterior  angles  is  equal  to  twice  as  many  right  angles 
as  the  polygon  has  sides.  Again,  the  sum  of  all  the  interior 
angles  is  equal  to  two  right  angles,  taken  as  many  times,  less 
two,  as  the  polygon  has  sides  (Prop.  XXVI.) ;  that  is,  equal  to 
twice  as  many  right  angles  as  the  figure  has  sides,  wanting 
four  right  angles.     Hence,  the  interior  angles  plus  four  right 


32  GEOMETRY. 

angles,  is  equal  to  twice 'as  many  right  angles  as  the  polygon 
has  oides,  and  consequently,  equal  to  the  sum  of  the  interior 
angles  plus  the  exterior  angles.  Taking  from  each  the  sum  of 
the  interior  angles,  and  there  remains  the  exterior  angles,  equal 
to  four  right  angles. 


PROPOSITION  XXVIII.    THEOREM.      . 

In  every  parallelogram,  the  opposite  sides  and  angles  are  equal. 

Let  ABCD  be  a  parallelogram :  then  will       jy c  » 

AB=DC,  AD=BC,  A=C,  and  ADC= ABC.        "'  ' 

For,  draw  the  diagonal  BD.  The  triangles 
ABD,  DBC,  have  a  common  side  BD ;  and 
since  AD,  BC,  are  parallel,  they  have  also  the 
angle  ADB=DBC,  (Prop.  XX.  Cor.  2.)  ;  and  since  AB,  CD, 
are  parallel,  the  angle  ABD==BDC  :  hence  the  two  triangles 
are  equal  (Prop.  VI.) ;  therefore  the  side  AB,  opposite  the  an- 
gle ADB,  is  equal  to  the  side  DC,  opposite  the  equal  angle 
DBC  ;  and  the  third  sides  AD,  BC,  are  equal:  hence  the  op- 
posite sides  of  a  parallelogram  are  equal. 

Again,  since  the  triangles  are  equal,  it  follows  that  the  angle 
A  is  equal  to  the  angle  C  ;  and  also  that  the  angle  ADC  com- 
posed of  the  two  ADB,  BDC,  is  equal  to  ABC,  composed  of 
the  two  equal  angles  DBC,  ABD  :  hence  the  opposite  apgles 
of  a  paKallelogram  are  also  equal. 

Cor.  Two  parallels  AB,  CD,  included  between  two  other 
parallels  AD,  BC,  are  equal ;  and  the  diagonal  DB  divides  the 
parallelogram  into  two  equal  triangles. 


PROPOSITION  XXIX.    THEOREM. 

Jf  the  opposite  sides  of  a  quadrilateral  are  equal,  each  to  each, 
the  equal  sides  will  he  parallel,  and  the  figure  will  he  a  par- 
allelogram. 

Let  ABCD  be  a  quadrilateral,  having 
its  opposite  sides  respectively  equal,  viz. 
AB=DC,  and  AD=:BC  ;  then  will  these 
sides  be  parallel,  and  the  figure  be  a  par- 
allelogram. 

For,  having  drawn  the  diagonal  BD, 
the  triangles  ABD,  BDC,  ha^Q  all  the  sides  of  the  one  equal  to 


BOOK  I.  33 

m 

Ihe  corresponding  sides  of  the  other ;  therefore  they  are  equal, 
and  the  angle  ADB,  opposite  the  side  AB,  is  equal  to  DBC, 
opposite  CD  (Prop.  X.) ;  therefore,  the  side  AD  is  parallel  to 
BC  (Prop.  XIX. Cor.  1.).  For  a  hke  reason  AB  is  parallel  to 
CD  :  therefore  the  quadrilateral  ABCD  is  a  parallelogram. 

PROPOSITION  XXX.    THEOREM. 

If  two  opposite  sides  of  a  quadrilateral  are  equal  and  parallel 
the  remaining  sides  will  also  he  equal  and  parallel,  and  the 
figure  will  he  a  parallelogram. 

Let  ABCD  be  a  quadrilateral,  having 
the  sides  AB,  CD,  equal  and  parallel ; 
then  will  the  figure  be  a  parallelogram. 

For,  draw  the  diagonal  DB,  dividing 
the  quadrilateral  into  two  triangles.  Then,  J- 
since  AB  is  parallel  to  DC,  the  alternate 
angles  ABD,  BDC,  are  equal  (Prop.  XX.  Cor.  2.) ;  moreover, 
the  side  DB  is  common,  and  the  side  AB=DC ;  hence  the  tri- 
angle ABD  is  equal  to  the  triangle  DBC  (Prop.  V.) ;  therefore, 
the  side  AD  is  equal  to  BC,  the  angle  ADB  =  DBC,  and  conse- 
quently AD  is  parallel  to  BC ;  hence  the  figure  ABCD  is  a 
parallelogram. 

PROPOSITION  XXXI.    THEOREM. 

The  two  diagonals  of  a  parallelogram  divide  each  other  into  equal 
parts,  or  mutually  bisect  each  other. 

Let  ABCD  be  a  parallelogram,  AC  and 
DB  its  diagonals,  intersecting  at  E,then  will 
AE=EC,  and  DE=EB. 

Comparing  the  triangles  ADE,  CEB,  we 
find  the  side  AD  =  CB  (Prop.  XXVIIL), 
the  angle  ADE=CBE,  and  the  angle 
DAE=ECB  (Prop.  XX.  Cor.  2.);  hence  those  triangles  are 
equal  (Prop.  VI.) ;  hence,  AE,  the  side  opposite  the  angle 
ADE,  is  equal  to  EC,  opposite  EBC ;  hence  also  DE  is  equal 
to  EB. 

Scholium.  In  the  case  of  the  rhombus,  the  sides  AB,  BC, 
being  equal,  the  triangles  AEB,  EBC,  have  all  the  sides  of  the 
one  equal  to  the  corresponding  sides  of  the  other,  and  are 
therefore  equal:  whence  it  follows  that  the  angles. AEB, BEC, 
are  equal,  and  therefore,  that  the  two  diagonals  of  a  rhombus 
cut  each  other  at  right  angles. 


34  GEOMETRY. 

BOOK  11. 

OF  RATIOS  AND  PROPORTIONS. 

Definitions, 

1.  Ratio  is  the  quotient  arising  from  dividing  one  quantity 
by  another  quantity  of  the  same  kind.  Thus,  if  A  and  B  rep- 
resent quantities  of  the  same  kind,  the  ratio  of  A  to  B  is  ex- 
pressed by  -V-. 

The  ratios  of  magnitudes  may  be  expressed  by  numbers, 
either  exactly  or  approximatively ;  and  in  the  latter  case,  the 
approximation  may  be  brought  nearer  to  the  true  ratio  than 
any  assignable  difference. 

Thus,  of  two  magnitudes,  one  of  them  may  be  considered  to 
be  divided  into  some  number  of  equal  parts,  each  of  the  same 
kind  as  the  whole,  and  one  of  those  parts  being  considered  as 
an  unit  of  measure,  the  magnitude  may  be  expressed  by  the 
number  of  units  it  contains.  If  the  other  magnitude  contain 
a  certain  number  of  those  units,  it  also  may  be  expressed  by 
the  number  of  its  units,  and  the  two  quantities  are  then  said 
to  be  commensurable. 

If  the  second  magnitude  do  not  contain  the  measuring  unit 
an  exact  number  of  times,  there  may  perhaps  be  a  smaller  unit 
which  will  be  contained  an  exact  number  of  times  in  each  of 
the  magnitudes.  But  if  there  is  no  unit  of  an  assignable  value, 
which  shall  be  contained  an  exact  number  of  times  in  each  of 
the  magnitudes,  the  magnitudes  are  said  to  be  incommensurable. 

It  is  plain,  however,  that  the  unit  of  measure,  repeated  as 
many  times  as  it  is  contained  in  the  second  magnitude,  would 
always  differ  from  the  second  magnitude  by  a  quantity  less 
than  the  unit  of  measure,  since  the  remainder  is  always  less 
than  the  divisor.  Now,  since  the  unit  of  measure  may  be  made 
as  small  as  we  please,  it  follows,  that  magnitudes  may  be  rep- 
resented by  numbers  to  any  degree  of  exactness,  or  they  will 
differ  from  their  numerical  representatives  by  less  than  any 
assignable  quantity. 

Therefore,  of  two  magnitudes,  A  and  B,  we  may  conceive 
A  to  be  divided  into  M  number  of  units,  each  equal  to  A' : 
then  A=M  x  A':  let  B  be  divided  into  N  number  of  equal  units, 
each  equal  to  A';  then  B=N  x  A';  M  and  N  being  integral  num- 
bers. Now  the  ratio  of  A  to  B,  will  be  the  same  as  the  ratio 
of  M  X  A'  to  N  X  A';  that  is  thasame  as  the  ratio  of  M  to  N,  since 
A'  is  a  common  unit. 


BOOK  11.  35 

In  the  same  manner,  the  ratio  of  any  other  two  magnitudes 
C  and  D  may  be  expressed  by  P  x  C  to  Q  x  C,  P  and  Q  being 
also  integral  numbers,  and  their  ratio  will  be  the  same  as  that 
ofPtoQ. 

2.  If  there  be  four  magnitudes  A,  B,  C,  and  D,  having  such 

values  that  —-is  equal  to— ,  then  A  is  said  to  have  the  same  ratio 
A  O 

to  B,  that  C  has  to  D,  or  the  ratio  of  A  to  B  is  equal  to  the  ratio 

of  C  to  D.     When  four  quantities  have  this  relation  to  each 

other,  they  are  said  to  be  in  proportion. 

To  indicate  that  the  ratio  of  A  to  B  is  equal  to  the  ratio  of 

C  to  D,  the  quantities  are  usually  written  thus,  A :  B  : :  C  :  D, 

and  read,  A  is  to  B  as  C  is  to  D.     The  quantities  which  are 

compared  together  are  called  the  terms  of  the  proportion.  The 

first  and  last  terms  are  called  the  two  extremes,  and  the  second 

and  third  terms^  the  two  means. 

3.  Of  four  proportional  quantities,  the  first  and  third  are 
called  the  antecedents,  and  the  second  and  fourth  the  conse- 
quents ;  and  the  last  is  said  to  be  a  fourth  proportional  to  the 
other  three  taken  in  order. 

4.  Three  quantities  are  in  proportion,  when  the  first  has  the 
same  ratio  to  the  second,  that  the  second  has  to  the  third ;  and 
then  the  middle  term  is  said  to  be  a  mean  proportional  between 
the  other  two. 

5.  Magnitudes  are  said  to  be  in  proportion  by  inversion,  or 
inversely,  when  the  consequents  are  taken  as  antecedents,  and 
the  antecedents  as  consequents. 

6.  Magnitudes  are  in  proportion  by  alternation,  or  alternately, 
vv^hen  antecedent  is  compared  with  antecedent,  and  consequent 
with  consequent. 

7.  Magnitudes  are  in  proportion  by  composition,  when  the 
sum  of  the  antecedent  and  consequent  is  compared  either  with 
antecedent  or  consequent. 

8.  Magnitudes  are  said  to  be  in  proportion  by  division,  when 
the  difference  of  the  antecedent  and  consequent  is  compared 
either  with  antecedent  or  consequent. 

9.  Equimultiples  of  two  quantities  are  the  products  which 
arise  from  multiplying  the  quantities  by  the  same  number : 
thus,  m  X  A,  m  X  B,  are  equimultiples  of  A  and  B,  the  common 
multiplier  being  m. 

10.  Two  quantities  A  and  B  are  said  to  be  reciprocally 
proportional,  or  inversely  proportional,  when  one  increases  in 
the  same  ratio  as  the  other  diminishes.  In  such  case,  either 
of  them  is  equal  to  a  constant  quantity  divided  by  the  other, 
and  their  product  is  constant. 


86  GEOMETRY. 


PROPOSITION  I.    THEOREM. 


k: 


When  four  quantities  are  in  proportion^  the  product  of  the 
two  extremes  is  equal  to  the  product  of  the  two  means 

Let  A,  B,  C,  D,  be  four  quantities  in  proportion,  and  M  :  N 

: ;  P  :  Q  be  their  numerical  representatives  ;  then  will  M  x  Q= 

N    O 
Nx  P;  for  since  the  quantities  are  in  proportion  — =-   there- 

O  MP 

fore  N=Mx^,orNxP=MxQ. 

Cor,  If  there  are  three  proportional  quantities  (Def.  4.),  the 
product  of  the  extremes  will  be  equal  to  the  square  of  the 
mean. 


PROPOSITION  II.    THEOREM. 

If  the  product  of  two  quantities  he  equal  to  the  pr-oduct  of  two  other 
quantities^  two  of  them  will  he  the  extremes  and  the  other  two 
the  means  of  a  proportion. 

Let  M  X  Q=N  X  P ;  then  will  M :  N  : :  P  :  Q. 

For,  if  P  have  not  to  Q  the  ratio  which  M  has  to  N,  let  P 
have  to  Q',  a  number  greater  or  less  than  Q,  the  same  ratio 
that  M  has  to  N;  that  is,  let  M  :  N  :  :  P  :  Q' ;  then  MxQ'= 

NxP  (Prop.  L)  :  hence,  Q'=  ^^^    ;  but  Q=Z^  ;  con- 


feequently,  Q=Q'  and  the  four  quantities  are  proportional;  that 
is,  M:N:;P:Q. 


PROPOSITION  III.    THEOREM. 

If  four  quantities  are  in  proportion,  they  will  he  in  proportion 
when  taken  alternately. 

Let  M,  N,  P,  Q,  be  the  numerical  representatives  of  four 
quanties  in  proportion  ;  so  that 

M  :  N  : :  P  :  Q,  then  will  M  :  P  : :  N  :  Q. 

Since  M  ;  N  : :  P  :  Q,  by  supposition,  M  x  Q=N  x  P  ;  there- 
fore, M  and  Q  may  be  made  the  extremes,  and  N  and  P  the 
means  of  a  proportion  (Prop.  IL) ;  hence,  M  :  P  : :  N  :  Q. 


BOOK  IT.  37 


PROPOSITION  IV. .  THEOREM. 

[f  there  he  four  proportional  quantities,  and  four  other  propor- 
tional quantities,  having  the  antecedents  the  same  in  both,  the 
consequefits  will  be  proportional. 

Let  M  :  N  : :  P  :  Q 

and  M  :  R  : :  P :  S 

then  will  N  :  Q  : :  R  :  S 

P     O 

For,  by  alternation      M;P::N:Q,or     tj=5- 

P      S 
and         M  :  P  ; :  R  ;  S,  or     5|=g" 

Q     S 
hence  :j^=:5-  ;  or  N  :  Q : :  R  :  S. 

Cor.  If  there  be  two  sets  of  proportionals,  having  an  ante- 
cedent and  consequent  of  the  first,  equal  to  an  antecedent  and 
consequent  of  the  second,  the  remaining  terms  will  be  propor- 
tional. 


,  PROPOSITION  V.    THEOREM. 

If  four  quantities  be  in  proportion,  they  will  be  in  proportion  when 
taken  inversely. 

Let  M:N::P:Q.;  then  will  t 

N  :  M  : :  Q  :  P. 
For,  from  the  first  proportion  we  have  M  x  Q=N  x  P,  or 
NxP=MxQ. 

But  the  products  N  x  P  and  M  x  Q  are  the  products  of  the 
extremes  and  mean^  of  the  four  quantities  N,  M ,  Q,  P,  and  these 
products  being  equal, 

N:M::Q:P(Prop.IL). 


PROPOSITION  VI.    THEOREM. 

If  four  quantities  are  in  proportion,  they  will  be  in  proportion  by 
composition,  or  division. 


D 


38  GEOMETRY. 

Let,  as  before,  M,  N,  P,  Q,  be  the  numerical  representatives 
of  the  four  quantities,  so  that 

M  :  N  : :  P  :  Q  ;  then  will 
M±N:M::Pd=Q:P. 
For,  from  the  first  proportion,  we  have 

MxQ=NxP,  orNxP=MxQ; 
Add  each  of  the  members  of  the  last  equation  to,  or  subtract 
It  from  M.P,  and  we  shall  have, 

M.P±N.P=:M.P±M.Q;or 
(M±N)xP-(P±Q)xM. 
But  MrbN  and  P,  maybe  considered  the  two  extremes,  and 
P±Q  and  M,  the  two  means  of  a  proportion :  hence, 
M±N  :  M  : :  FiQ  :  P. 


PROPOSITION  VII.    THEOREM. 

Equimultiples  of  any  two  quantities^  have  the  same  ratio  as  the 
quantities  themselves. 

Let  M  and  N  be  any  two  quantities,  and  m  any  integral 
number ;  then  will 

m.M.:m.  N  : :  M  :  N.     For 

m.  MxN=?w.  NxM,  since  the  quantities  in 
each  member  are  the  same ;  therefore,  the  quantities  are  pro- 
portional (Prop.  II.) ;  or 

m.  M  :  771.  N  : :  M  :  N. 


PROPOSITION  VIII.    THEOREM. 

Of  four  proportional  quantities,  if  there  he  taken  any  equimul- 
tiples of  the  two  antecedents,  and  any  equimultiples  of  the  two 
consequents,  the  four  resulting  quantities  will  be  proportional 

Let  M,  N,  P,  Q,  be  the  numerical  representatives  of  four 
quantities  in  proportion ;  and  let  m  and  n  be  any  numbers 
whatever,  then  will 

m.  M  :  ?i.  N  : :  m.  P  :  71.  Q. 

For,  since  M  :  N  : :  P  :  Q,  we  have  M  x  Q=N  x  P ;  hence, 
m.  M  X  71.  Q=?i.  N  X  77Z.  P,  by  multiplying  both  members  of  the 
equation  by  ??i  x  n.  But  m,  M  and  n.  Q,  may  be  regarded  as 
the  two  extremes,  and  n.  N  and  m.  P,  as  the  means  of  a  propor- 
tion ;  hence,  tw.  M  :  n.  N : ;  m,  P  :  ti.  Q. 


BOOK  II.  a^ 


PROPOSITION  IX.    THEOREM. 

Of  four  proportional  quantities,  if  the  two  consequents  he  either 
augmented  or  diminished  hy  quantities  which  have  the  same 
ratio  as  the  antecedents,  the  resulting  quantities  and  the  ante- 
cedents will  he  proportional. 


Let 

M  :  N  :  :  P  :  Q,  and  let  also 

M  :  P  :  :  m  :  71,  then  will 

M  :  P  :  :  N±»z  :  Qrbw. 

For,  since 

M  :  N  :  :  P  :  Q,  MxQ=NxP. 

And  since 

M  :  P  :  :  m  :  71,  Mx7i=Px7?i 

Therefore, 

MxQ±Mx7i=NxP±Px77z 

or. 

Mx(Q=i=7i)=Px(N±m): 

hence 

M  :  P  :  :  Ndbm  :  Q±7i  (Prop.  II.). 

PROPOSITION  X.    THEOREM. 

[f  any  numher  of  quantities  are  proportionals,  any  one  antece- 
dent will  he  to  its  consequent,  as  the  sum  of  all  the  antecedents 
to  the  sum  of  the  consequents. 

Let  M  :  N  :  ;  P  :  Q  :  :  R  j^  &c^then  will 

M  :  N  :  :  M  +  P  +  R  :  N  +  Q  +  S 
For,  since        M  :  N  :  :  P  :  Q,  we  have  MxQ=JixP 
And  since        M  :  N  :  :  R  :  S,  we  have  Mx  S=]?lxR 

Add  MxN=MxN 

and  we  have,    M.N+M.Q  +  M.S=M.N+N.P  +  N.R 
or     Mx(N  +  Q+S)^Nx(M  +  P  +  R) 
therefore,  M  :  N  :  :  M+P  +  R  :  N  +  Q+S. 


PROPOSITION  XI.    THEOREM. 

If  two  magnitudes  he  each  increased  or  diminished  hy  like  parts 
of  each,  the  resulting  quantities  will  have  the  same  ratio  as  the 
magnitudes  themselves. 


m 


40  GEOMETRY. 

Let  M  and  N  be  any  two  magnitudes,  and —  and  i-  be  like 

m         m 

parts  of  each  :  then  will 

M  ;  N  :  :  Md=M  :  N  ±— 
m  m 

For,  it  is  obvious  that  Mx(N±_\  =Nx(M±_\  since 

m  /  m  / 

each   is  equal  to  M.Ndb_L_.     Consequently,  the  four  quan- 

m 
titles  are  proportional  (Prop.  II.), 


PROPOSITION  XII.    THEOREM. 

//*  four  quantities  are  proportional,  their  squares  or  cubes  will 
also  be  proportional. 

Let  M  :  N  :  P  :  Q, 

then  will  M2  :  N2  :  :  P2  :  Q2 

and  M^  :  N^  :  :  P3  :  Q3 

For,      MxQ=NxP,  since  M  :  N  :  :  P  :  Q 

or,      M^  X  Q^=N^  X  P^      by  squaring  both  members, 
and      M^  X  Q^=N^  X  P^      by  cubing  both  members ; 
therefore,     M^  ;  N^  :  :  P^  :  Q2 
and      M^  :  N^  :  :  P3  :  Q3 

Cor.  In  the  same  way  it  may  be  shown  that  like  powers  or 
roots  Jtf  proportional  quantities  are  proportionals. 

PROPOSITION  XIII.    THEOREM. 

Jf  there  be  two  sets  of  proportional  quantities,  the  products  ofths 
corresponding  terms  will  be  proportional. 


Let 
and 
then  will 
For  since 
and 

M  :  N  :  :  P  :  Q 
.  R  •  S  :  :  T  :  V 
*MxR  :  NxS  :  :  PxT  :  QxV 

MxQ=N>?P 

R  X  V=  S  X  T,  we  shall  have 

MxQx^xV=NxPxSxT 

or 

MxRxQxV=NxSxPxT 

therefore, 

MxR  :  NxS  :  :  PxT  :  QxV 

BOOK  III.  41 

BOOK  III. 

THE  CIRCLE,  AND  THE  MEASUREMENT  OF  ANGLES. 

Definitions, 

1.  The  circumference  of  a  circle  is  a 
curved  line,  all  the  points  of  which  are 
equally  distant  from  a  point  within, 
called  the  centre. 

The  circle  is  the  space  terminated  by  ^.j 
this  curved  line.* 

2.  Every  straight  line,  CA,  CE,  CD, 
drawn  from  the  centre  to  the  circum- 
ference, is  called  a  radium  or  semidiam-  E" 
eter ;  every  line  which,  like  AB,  passes  through  the  centre,  and 
is  terminated  on  both  sides  by  the  circumference,  is  called  a 
diameter. 

From  the  definition  of  a  circle,  it  follows  that  all  the  radii 
are  equal ;  that  all  the  diameters  are  fqual  also,  and  each 
double  of  the  radius. 

3.  A  portion  of  the  circumference,  such  as  FHG,  is  called 
an  arc. 

The  chord,  or  subtense  of  an  arc,  is  the  straight  line  FG,  which 
joins  its  two  extremities.f 

4.  A  segment  is  the  surface  of  portion  of  a  circle,  included 
between  an  arc  and  its  chord. 

5.  A  sector  is  the  part  of  the  circle  included  between  an 
arc  DE,  and  the  two  radii  CD,  CE,  drawn  to  the  extremities 
of  the  arc. 

6.  A  straight  line  is  said  to  be  inscribed  in 
a  circle,  when  its  extremities  are  in  the  cir- 
cumference, as  AB. 

An  inscribed  angle  is  one  which,  like  BAC, 
has  its  vertex  in  the  circumference,  and  is 
formed  by  two  chords. 


*  Note.  In  common  language,  the  circle  is  sometimes  confounded  with  its 
circumference :  but  the  correct  expression  may  always  be  easily  recurred  to  if 
we  bear  in  mind  that  the  circle  is  a  surface  which  has  length  and  breadth, 
while  the  circumference  is  but  a  line. 

t  Note.     In  all  cases,  the  same  chord  FG  belongs  to  two  arcs,  FGH,  FEG, 
and  consequently  also  to  two  segments  :  but  the  smaller  one  is  always  meant, 
#      unless  the  contrary  is  expressed. 

D* 


42  GEOMETRY. 

An  inscribed  triangle  is  one  which,  like  BAG,  has  its  three 
angular  points  in  the  circumference. 

And,  generally,  an  inscribed  figure  is  one,  of  which  all  the 
angles  have  their  vertices  in  the  circumference.  The  circle  is 
then  said  to  circumscribe  such  a  figure. 

7.  A  secant  is  a  line  which  meets  the  circum- 
ference  in  two  points,  and  lies  partly  within  ^- 
and  partly  without  the  circle.   AB  is  a  secant. 

8.  A  tangent  is  a  hne  which  has  but  one 
point  in  common  with  the  circumference.  CD 
is  a  tangent. 

The  point  M,  where  the  tangent  touches  the  c 
circumference,  is  called  the  point  of  contact. 


In  like  manner,  two  circumferences  touch 
each  other  when  they  have  but  one  point  in 
common. 


9.  A  polygon  is  circumscribed  about  a 
circle,  when  ail  its  sides  are  tangents  to 
the  circumference  :  in  the  same  case,  the 
circle  is  said  to  be  inscHbed  in  the  po- 
lygon. 


PROPOSITION  I.    THEOREM. 

Every  diameter  divides  the  circle  and  its  circumference  into  two 
equal  parts. 

Let  AEDF  be  a  circle,  and  AB  a  diameter. 
Now,  if  the  figure  AEB  be  applied  to  AFB, 
their  common  base  AB  retaining  its  position, 
the  curve  line  AEB  must  fall  exactly  on  the 
curve  line  AFB,  otherwise  there  would,  in 
the  one  or  the  other,  be  points  unequally  dis- 
tant from  the  centre,  which  is  contrary  to 
tlie  definition  of  a  circle.     ^ 


I 


BOOK  III.  43 


PROPOSITION  II.    THEOREM. 
Every  chord  is  less  than  the  diameter. 


Let  AD  be  any  chord.     Draw  the  radii 
CA,  CD,  to  its  extremities.     We  shall  then 
have  AD<AC  +  CD  (Book  I.  Prop.  VII.*);  ^ 
or  AD<Ap. 


Cor,  [Hence  the  greatest  Hne  which  can  be  inscribed  in  a 
Y     circle  is  its  diameter.^ 


PROPOSITION  III.    THEOREM. 

A  straight  line  cannot  meet  the  circumference  of  a  circle  in  more 
than  two  points. 

For,  if  it  could  meet  it  in  three,  those  three  points  would  be 
equally  distant  from  the  centre  ;  and  hence,  there  would  be 
three  equal  straight  lines  drawn  from  the  same  point  to  the 
same  straight  line,  which  is  impossible  (Book  I.  Prop.  XV. 
Cor.  2.). 


PROPOSITION  IV.    THEOREM. 

In  the  same  circle,  or  in  equal  circles,  equal  arcs  are  subtended  by 
equal  chords ;  and,  conversely,  equal  chords  subtend  equal  arcs. 


Note.  When  reference  is  made  from  one  pronosition  to  another,  in  the 
same  Book,  the  number  of  the  proposition  referred  to  is  alone  given;  but  when 
the  proposition  is  found  in  a  different  Book,  the  number  of  the  Book  is  also 
given. 


44  GEOMETRY. 

If  the  radii  AC,  EO,  are 
equal,    and    also    the    arcs 
AMD,  ENG;  then  the  chord 
AD  will  be  equal    to   the  jA 
chord  EG. 

For,  since  the  diameters 
AB,  EF,  are  equal,  the  semi- 
circle AMDB  maybe  applied 
exactly  to  the  semicircle  ENGF,  and  the  curve  line  AMDB 
will  coincide  entirely  with  the  curve  line  ENGF.  But  the 
part  AMD  is  equal  to  the  part  ENG,  by  hypothesis ;  hence,  the 
point  D  will  fall  on  G  ;  therefore,  the  chord  AD  is  equal  to  the 
chord  EG. 

Conversely,  supposing  again  the  radii  AC,  EO,  to  be  equal, 
if  the  chord  AD  is  equal  to  the  chord  EG,  the  arcs  AMD, 
ENG  will  also  be  equal. 

For,  if  the  radii  CD,  OG,  be  drawn,  the  triangles  ACD, 
EOG,  will  have  all  their  sides  equal,  each  to  each,  namely, 
AC^EO,  CD  =  OG,  and  AD=EG;  hence  the  triangles  are 
themselves  equal ;  and,  consequently,  the  angle  ACD  is  equal 
EOG  (Book  I.  Prop.  X.).  Now,  placing  the  semicircle  ADB 
on  its  equal  EGF,  since  the  angles  ACD,  EOG,  are  equal,  it  is 
plain  that  the  radius  CD  will  fall  on  the  radius  OG,  and  the 
point  D  on  the  point  G  ;  therefore  the  arc  AMD  is  equal  to  the 
arc  ENG. 


PROPOSITION  V.    THEOREM. 

In  the  same  circle y  or  in  equal  circles ,  a  greater  arc  is  subtended 
by  a  greater  chords  and  conver'sely^  the  greater  chord  subtend 
the  greater  arc. 

Let  the  arc  AH  be  greater  than 
the  arc  AD  ;  then  will  the  chord  AH 
be  greater  than  the  chord  AD. 

For,  draw  the  radii  CD,  CH.  The 
two  sides  AC,  CH,  of  the  triangle   .  / 
ACH  are  equal  to  the  two  AC,  CD,  -^^ 
of  the  triangle  ACD,  and  the  angle 
ACH  is  greater  than  ACD ;  hence,  the 

third  side  AH  is  greater  than  the  third  

side  AD  (Book  I.  Prop.  IX.)  ;  there-  X 

fore  the  chord,  which  subtends  the  greater  arc,  is  the  greater. 
Conversely,  if  the  chord  AH  is  greater  than  AD,  it  will  follow, 
on  comparing  the  same  triangles,  that  the  angle  ACH   is 


BOOK  III.  4i, 

greater  than  ACD  (Bk.  I.  Prop.  IX.  Sch.) ;  and  hence  that 
the  arc  AH  is  greater  than  AD ;  since  the  whole  is  greater 
than  its  part. 

Scholium.  The  arcs  here  treated  of  are  each  less  than  the 
semicircumference.  If  they  were  greater,  the  reverse  pro- 
perty would  have  place  ;  for,  as  the  arcs  increase,  the  chords 
would  diminish,  and  conversely.  Thus,  the  arc  AKBD  is 
greater  than  AKBH,  and  the  chord  AD,  of  the  first,  is  less 
than  the  chord  AH  of  the  second. 


PROPOSITION  VI.    THEOREM. 

The  radius  which  is  perpendicular  to  a  chord,  bisects  the  chords 
and  bisects  also  the  subtended  arc  of  the  chord. 

Let  AB  be  a  chord,  and  CG  the  ra- 
dius perpendicular  to  it :  then  will  AD  = 
DB,  and  the  arc  AG^GB. 

For,  draw  the  radii  CA,  CB.  Then 
the  two  right  angled  triangles  ADC, 
CDB,  will  have  AC=CB,  and  CD  com- 
mon ;  hence,  AD  is  equal  to  DB  (Book 
T.  Prop.  XVII.). 

Again,  since  AD,  DB,  are  equal,  CG 
is  a  perpendicular  erected  from  the  mid- 
dle of  AB  ;  hence  every  point  of  this  perpendicular  must  be 
equally  distant  from  its  two  extremities  A  and  B  (Book  I.  Prop. 
XVI.).  Now,  G  is  one  of  these  points  ;  therefore  AG,  BG,  are 
equal.  But  if  the  chord  AG  is  equal  to  the  chord  GB,  the  arc 
AG  will  be  equal  to  the  arc  GB  (Prop.  IV.)  ;  hence,  the  radius 
CG,  at  right  angles  to  the  chord  AB,  divides  the  arc  subtended 
by  that  chord  into  two  equal  parts  at  the  point  G. 

Scholium.  The  centre  C,  the  middle  point  D,  of  the  chord 
AB,  and  the  middle  point  G,  of  the  arc  subtended  by  this 
chord,  are  three  points  of  the  same  line  perpendicular  to  the 
chord.  But  two  points  are  sufficient  to  determine  the  position 
of  a  straight  line ;  hence  every  straight  line  which  passes  through 
two  of  the  points  just  mentioned,  will  necessarily  pass  through 
the  third,  and  be  perpendicular  to  the  chord. 

It  follows,  likewise,  that  the  perpendicular  raised  from  the 
middle  of  a  chord  passes  through  the  centre  of  the  circle y  and 
through  the  middle  of  the  arc  subtended  by  that  chord. 

For,  this  perpendicular  is  the  same  as  the  one  let  fall  from 
the  centre  on  the  same  chord,  since  both  of  them  pass  through 
the  centre  and  middle  of  the  chord. 


46  GEOMETRY. 


PROPOSITION  VII.    THEOREM. 


fi-^<i^^jf^\ 


Through  three  given  points  not  in  the  same  straight  line,  one  cir- 
cumference may  always  he  made  to  pass,  and  but  one. 

Let  A,  B,  and  C,  be  the  given 
points. 

Draw  AB,  BC,  and  bisect  these 
straight  Unes  by  the  perpendiculars 
DE,  FG :  we  say  first,  that  DE  and 
FG,  will  meet  in  some  point  O. 

For,  they  must  necessarily  cut 
each  other,  if  they  are  not  parallel. 
Now,  if  they  were  parallel,  the  line  AB,  which  is  perpendicular 
to  DE,  would  also  be  perpendicular  to  FG,  and  the  angle  K 
would  be  a  right  angle  (Book  I.  Prop.  XX.  Cor.  1.).  But  BK, 
the  prolongation  of  BD,  is  a  difierent  line  from  BF,  because  the 
three  points  A,  B,  C,  are  not  in  the  same  straight  line  ;  hence 
there  would  be  two  perpendiculars,  BF,  BK,  let  fall  from  the 
same  point  B,  on  the  same  straight  line,  which  is  impossible 
(Book  I.  Prop.  XIV.)  ;  hence  DE,  FG,  will  always  meet  in 
some  point  O. 

And  moreover,  this  point  O,  since  it  lies  in  the  perpendicular 
DE,  is  equally  distant  from  the  two  points,  A  and  B  (Book  I. 
Prop.  XVI.)  ;  and  since  the  same  point  O  lies  in  the  perpen- 
dicular FG,  it  is  also  equally  distant  from  the  two  points  B  and 
C  :  hence  the  three  distances  OA,  OB,  OC,  are  equal ;  there- 
fore the  circumference  described  from  the  centre  O,  with  the 
radius  OB,  will  pass  through  the  three  given  points  A,  B,  C. 

We  have  now  shown  that  one  circumference  can  always  be 
made  to  pass  through  three  given  points,  not  in  the  same 
straight  line  :  we  say  farther,  that  but  one  can  be  described 
through  them. 

For,  if  there  were  a  second  circumference  passing  through  the 
three  given  points  A,  B,  C,  its  centre  could  not  be  out  of  the 
line  DE,  for  then  it  would  be  unequally  distant  from  A  and  B 
(Book  I.  Prop.  XVI.);  neither  could  it  be  out  of  the  line  FG,  for 
a  like  reason  ;  therefore,  it  would  be  in  both  the  lines  DE,  FG. 
But  two  straight  lines  cannot  cut  each  other  in  more  than  one 
point ;  hence  there  is  but  one  circumference  which  can  pass 
through  three  given  points. 

Cor.  Two  circumferences  cannot  meet  in  more  than  two 
pomts  ;  for,  if  they  have  three  common  points,  there  would  be 
two  circumferences  passing  through  the  same  three  points ; 
which  has  been  shown  by  the  proposition  to  be  impossible. 


BOOK  III. 


47 


\. 


■\ 


PROPOSITION  VIII.    THEOREM. 


Two  equal  chords  are  equally  distant  from  the  centre  ;  andoj  two 
unequal  chords,  the  less  is  at  the  greater  distance  from  the 
centre. 


First.  Suppose  the  chord  AB= 
DE.  Bisect  these  chords  by  the  per- 
pendiculars CF,  CG,  and  draw  the 
radii  CA,  CD. 

In  the  right  angled  triangles  CAF, 
DCG,  the  hypothenuses  CA,  CD,  are 
equal ;  and  the  side  AF,  the  h^lf  of 
AB,  is  equal  to  the  side  DG,  the  half 
of  DE :  hence  the  triangles  are  equal, 
and  CFis  eciual  to  CG  (Book  I.  Prop. 
XVII.)  ;  hence,  the  two  equal  chords 
AB,  DE,  are  equally  distant  from  the  centre. 

Secondly  Let  the  chord  AH  be  greater  than  DE.  The 
arc  AKH  will  be  greater  than  DME  (Prop.  V.) :  cut  off  from 
the  former,  a  part  ANB,  equal  to  DME  ;  draw  the  chord  AB, 
and  let  fall  CF  perpendicular  to  this  chord,  and  CI  perpendicu- 
lar to  AH.  It  is  evident  that  CF  is  greater  than  CO,  and  CO 
than  CI  (Book  I.  Prop.  XV.)  ;  therefore,  CF  is  still  greater 
than  CI.  But  CF  is  equal  to  CG,  because  the  chords  AB, 
DE,  are  equal :  hence  we  have  CG>CI ;  hence  of  two  unequal 
chords,  the  less  is  the  farther  from  the  centre. 


n^^ 


PROPOSITION  IX.    THEOREM. 


A  straight  line  perpendicular  to  a  radius,  at  its  extremity,  is  a 
tangent  to  the  circumference. 

Let  BD  be  perpendicular  to  the  B 
radius  C  A,  at  its  extremity  A ;  then 
will  it  be  tangent  to  the  circumfe- 
rence. 

For,  every  oblique  line  CE,  is 
longer  than  the  perpendicular  CA 
(Book  I.  Prop.  XV.);  hence  the 
point  E  is  without  the  circle  ;  therefore,  BD  has  no  point  but 
A  common  to  it  and  the  circumference  ;  consequently  BD  is  a 
tangent  (Def.  8.). 


48 


GEOMETRY. 


Scholium.  At  a  given  point  A,  only  one  tangent  AD  can 
be  drawn  to  the  circumference  ;  for,  if  another  could  be  drawn, 
it  would  not  be  perpendicular  to  the  radius  CA  (Book  I.  Prop. 
XIV.  Sch.)  ;  hence  in  reference  to  this  new  tangent,  the  radius 
AC  would  be  an  oblique  line,  and  the  perpendicular  let  fall 
from  the  centre  upon  this  tangent  would  be  shorter  than  CA ; 
hence  this  supposed  tangent  would  enter  the  circle,  and  be  a 
secant. 


PROPOSITION  X.    THEOREM. 


aTm: 


Two  parallels  intercept  equal  arcs  on  the  circumference. 

There  may  be  three  cases. 

First.  If  the  two  parallels  are  se- 
cants, draw  the  radius  CH  perpendicu- 
lar to  the  chord  MP.  It  will,  at  the 
same  time  be  perpendicular  to  NQ 
(Book  I.Prop.XX.Cor.  1 .) ;  therefore,  the 
point  H  will  be  at  once  the  middle  of 
the  arc  MHP,  and  of  the  arc  NHQ 
(Prop.  VI.) ;  therefore,  we  shall  have 
the  arc  MH=HP,  and  the  arc  NH  = 
HQ  ;  and  therefore  MH— NH=:HP— HQ  ;  in  other  words, 
MN=PQ. 

Second.  When,  of  the  two  paral- 
lels AB,  DE,  one  is  a  secant,  the 
other  a  tangent,  draw  the  radius  CH 
to  the  point  of  contact  H  ;  it  will  be 
perpendicular  to  the  tangent  DE 
(Prop.  IX.),  and  also  to  its  parallel 
MP.  But,  since  CH  is  perpendicular 
to  the  chord  MP,  the  point  H  must  be 
the  middle  of  the  arc  MHP  (Prop. 
VI.) ;  therefore  the  arcs  MH,  HP,  in- 
cluded between  the  parallels  AB,  DE,  are  equal. 

Third.  If  the  two  parallels  DE,  IL,  are  tangents,  the  one 
at  H,  the  other  at  K,  draw  the  parallel  secant  AB  ;  and,  from 
what  has  just  been  shown,  we  shall  have  MH=HP,  MK=KP; 
and  hence  the  whole  arc  HMK=HPK.  It  is  farther  evident 
that  each  of  these  arcs  is  a  semicircumference. 


BOOK  III. 


49 


PROPOSITION  XI.    THEOREM. 

If  two  circles  cut  each  other  in  two  points,  the  line  which  passes 
through  their  centres,  will  be  perpendicular  to  the  chord  which 
joins  the  points  of  intersection,  and  will  divide  it  into  two 
equal  parts. 

For,  let  the  line  AB  join  the  points  of  intersection.     It  will 
be  a  common  chord  to  the  two  circles.  Now  if  a  perpendicular 


be  erected  from  the  middle  of  this  chord,  it  will  pass  through 
each  of  the  two  centres  C  and  D  (Prop.  VI.  Sch.).  But  no 
more  than  one  straight  line  can  be  drawn  through  two  points  ; 
hence  the  straight  line,  which  passes  through  the  centres,  will 
bisect  the  chord  at  right  angles. 


PROPOSITION  Xn.    THEOREM. 


If  the  distance  between  the  centres  of  two  circles  is  less  than  the 
sum  of  the  radii,  the  greater  radius  being  at  the  same  time 
less  than  the  sum  of  the  smaller  and  the  distance  between  the 
centres,  the  two  circumference^'jivill  cut  each  other. 

For,  to  make  an  intersection 
possible,  the  triangle  CAD  must 
be  possible.  Hence,  not  only- 
must  we  have  CD  <  AC  +  AD, 
but  also  the  greater  radius  AD< 
AC  +  CD  (Book  I.  Prop.  VII.). 
And,  whenever  the  triangle  CAD 
can  be  constructed,  it  is  plain 
that  the  circles  described  from  the  centres  C  and  D,  will  cut 
each  other  in  A  and  B. 


m 


50 


GEOMETRY. 


PROPOSITION  XIII.    THEOREM. 


If  the  distance  between  the  centres  of  two  circles  is  equal  to  the 
sum  of  their  radii,  the  two  circles  will  touch  each  other  exter- 
nally. 


Let  C  and  D  be  the  centres  at  a 
distance  from  each  other  equal  to 
CA  +  AD. 

The  circles  will  evidently  have  the 
point  A  common,  and  they  v^ill  have 
no  other;  because,  if  they  had  two 
points  common,  the  distance  between 


their  centres  must  be  less  than  the  sum  of  their  radii. 


PROPOSITION  XIV.    THEOREM. 


)( 


If  ike  distance  between  the  centres  of  two  circles  is  equal  to  the 
difference  of  their  radii,  the  two  circles  icill  touch  each  other 
internally. 

Let  C  and  D  be  the  centres  at  a  dis- 
tance from  each  other  equal  to  AD — CA. 

It  is  evident,  as  before,  that  they  will 
have  the  point  A  common :  they  can  have 
no  other;  because,  if  they  had,  the  greater 
radius  AD  must  be  less  than  the  sum  of 
the  radius  AC  and  the  distanceCD  between 
the  centres  (Prop.  XIL);  which  is  contraiy 
to  the  supposition. 

Cor.  Hence,  if  two  circles  touch  each  othef,  either  eiter- 
nally  or  internally,  their  centres  and  the  point  of  contact  "S^ill 
be  in  the  same  right  line. 

Scholium.  All  circles  which  have  their  centres  on  the  right 
line  AD.  and  which  pass  through  the  point  A,  are  tangent  to 
each  other.  For,  they  have  only  the  point  A  common,  and  if 
through  the  point  A,  AE  be  drawn  perpendicular  to  AD,  the 
straight  line  AE  will  be  a  common  tangent  to  all  the  circles. 


BOOK  III.  51 


PROPOSITION  XV.    THEOREM. 


KW^ 


In  the  same  circle,  or  in  equal  circles,  equal  angles  having  their 
vertices  at  the  centre,  intercept  equal  arcs  on  the  circumference  : 
and  conversely,  if  the  arcs  intercepted  are  equal,  the  angles 
contained  by  the  radii  will  also  be  equal,  • 

Let  C  and  C  be  the  centres  of  equal  circles,  and  the  angle 
ACB-DCE. 

First.  Since  the  angles  ACB, 
DCE,  are  equal,  they  may  be 

placed  upon  each   other ;    and   [         C         \  (         C 
since  their  sides  are  equal,  the 
point  A  will  evidently  fall  on  D, 

and  the  point  B  on  E.     But,  in    iiN J^B     3^ 

that  case,  the  arc  AB  must  also 

fall  on  the  arc  DE  ;  for  if  the  arcs  did  not  exactly  coincide,  there 
would,  in  the  one  or  the  other,  be  points  unequally  distant  from 
the  centre  ;  which  is  impossible  :  hence  the  arc  AB  is  equal 
to  DE. 

Secondly.  If  we  suppose  AB=DE,  the  angle  ACB  will  be 
equal  to  DCE.  For,  if  these  angles  are  not  equal,  suppose 
ACB  to  be  the  greater,  and  let  ACI  be  taken  equal  to  DCE. 
From  what  has  just  been  shown,  we  shall  have  AI=DE  :  but, 
by  hypothesis,  AB  is  equal  to  DE ;  hence  AI  must  be  equal  to 
AB,  or  a  part  to  the  whole,  which  is  absurd  (Ax.  8.)  :  hence, 
the  angle  ACB  is  equal  to  DCE. 


PROPOSITION  XVI.    THEOREM. 

In  the  same  circle,  or  in  equal  circles,  if  two  angles  at  the  centre 
are  to  each  other  in  the  proportion  of  two  whole  numbers,  the 
intercepted  arcs  will  be  to  each  other  in  the  proportion  of  the 
same  numbers,  and  we  shall  have  the  angle  to  the  angle,  as  the 
corresponding  arc  to  the  corresponding  arc. 


62 


GEOMETRY. 


Suppose,  for  example,  that  the  angles  ACB,  DCE,  are  to 
each  other  as  7  is  to  4;  or,  which  is  the  same  thing,  suppose 
that  the  angle  M,  which  may  serve  as  a  common  measure,  is 
contained  7  times  in  the  angle  ACB,  and  4  times  in  DCE 

C  C 


The  seven  partial  angles  ACm,  mCUf  nCp,  &c.,  into  which 
ACB  is  divided,  being  each  equal  to  any  of  the  four  partial 
angles  into  which  DCE  is  divided  ;  each  of  the  partial  arcs 
Am,  mn,  np,  <fec.,  will  be  equal  to  each  of  the  partial  arcs  Da-, 
xy,  &c.  (Prop.  XV.).  Therefore  the  whole  arc  AB  will  be  to 
the  whole  arc  DE,  as  7  is  to  4.  But  the  same  reasoning  would 
evidently  apply,  if  in  place  of  7  and  4  any  numbers  whatever 
were  employed ;  hence,  if  the  ratio  of  the  angles  ACB,  DCE, 
can  be  expressed  in  whole  numbers,  the  arcs  AB,  DE,  will  be 
to  each  other  as  the  angles  ACB,  DCE. 

Scholium,  Conversely,  if  the  arcs,  AB,  DE,  are  to  each 
other  as  two  whole  numbers,  the  angles  ACB,  DCE  will  be  to 
each  other  as  the  same  whole  numbers,  and  we  shall  have 
ACB  :  DCE  :  :  AB  :  DE.  For  the  partial  arcs.  Am,  7W7i,&c. 
and  Dx,  xy,  &c.,  being  equal,  the  partial  angles  ACm,  mCn, 
&c.  and  DCa;,  xCy,  &c.  will  also  be  equal. 


PROPOSITION  XVII.    THEOREM. 

Whatever  be  the  ratio  of  two  angles,  they  will  always  he  to  each 
other  as  the  arcs  intercepted  between  their  sides ;  the  arcs  being 
described  from  the  vertices  of  the  angles  as  centres  with  equal 
radii. 


Let  ACB  be  the  greater  and 
ACD  the  less  angle. 

Let  the  less  angle  be  placed 
on  the  greater.  If  the  propo- 
sition is  not  true,  the  angle 
ACB  will  be  to  the  angle  ACD 
as  the  arc  AB  is  to  an  arc 
greater  *or  less  than  AD.  Suppose  this  arc  to  be  greater,  and 
let  it  be  represented  by  AO  ;  we  shall  thus  have,  the  angle 
ACB  :  angle  ACD  : :  arc  AB  :  arc  AO.  Next  conceive  the  ai'c 


BOOK  III.  53 

AB  to  be  divided  into  equal  parts,  each  of  which  is  less  than 
DO  ;  there  will  be  at  least  one  point  of  division  between  D  and 
O  ;  let  I  be  that  point ;  and  draw  CI.  The  arcs  AB,  AI,  will  be 
to  each  other  as  two  whole  numbers,  and  by  the  preceding 
theorem,  we  shall  have,  the  angle  ACB  :  angle  ACI :  :  arc  AB 
:  arc  AI.  Comparing  these  two  proportions  with  each  other, 
we  see  that  the  antecedents  are  the  same  :  hence,  the  conse- 
quents are  proportional  (Book  II.  Prop.  IV.) ;  and  thus  we  find 
the  angle  ACD  :  angle  ACI : :  arc  AO  :  arc  AI.  But  the  arc 
AO  is  greater  than  the  arc  AI ;  hence,  if  this  proportion  is  true, 
the  angle  ACD  must  be  greater  than  the  angle  ACI  :  on  the 
contrary,  however,  it  is  less ;  hence  the  angle  ACB  cannot  be 
to  the  angle  ACD  as  the  arc  AB  is  to  an  arc  greater  than  AD. 
By  a  process  of  reasoning  entirely  similar,  it  may  be  shown 
that  the  fourth  term  of  the  proportion  cannot  be  less  than  AD  ; 
hence  it  is  AD  itself ;  therefore  we  have 

Angle  ACB  :  angle  ACD  :  :  arc  AB  :  arc  AD. 

Cor.  Since  the  angle  at  the  centre  of  a  circle,  and  the  arc 
intercepted  by  its  sides,  have  such  a  connexion,  that  if  the  one 
be  augmented  or  diminished  in  any  ratio,  the  other  will  be 
augmented  or  diminished  in  the  same  ratio,  we  are  authorized 
to  establish  the  one  of  those  magnitudes  as  the  measure  of  the 
other ;  and  we  shall  henceforth  assume  the  arc  AB  as  the  mea- 
sure of  the  angle  ACB.  It  is  only  necessary  that,  in  the  com- 
parison of  angles  with  each  other,  the  arcs  which  serve  to 
measure  them,  be  described  with  equal  radii,  as  is  implied  in 
all  the  foregoing  propositions. 

Scholium  1.  It  appears  most  natural  to  measure  a  quantity 
by  a  quantity  of  the  same  species  ;  and  upon  this  principle  it 
would  be  convenient  to  refer  all  angles  to  the  right  angle  ; 
which,  being  made  the  unit  of  measure,  an  acute  angle  would 
be  expressed  by  some  number  between  0  and  1 ;  an  obtuse  an- 
gle by  some  number  between  1  and  2.  This  mode  of  express- 
ing angles  would  not,  how^ever,  be  the  most  convenient  in 
practice.  It  has  been  found  more  simple  to  measure  them  by 
arcs  of  a  circle,  on  account  of  the  facility  with  which  arcs  can 
be  made  equal  to  given  arcs,  and  for  various  other  reasons.  At 
all  events,  if  the  measurement  of  angles  by  arcs  of  a  circle 
is  in  any  degree  indirect,  it  is  still  equally  easy  to  obtain  the 
direct  and  absolute  measure  by  this  method ;  since,  on 
comparing  the  arc  which  serves  as  a  measure  to  any  an- 
gle, with  the  fourth  part  of  the  circumference,  we  find  the' 
ratio  of  the  given  angle  to  a  right  angle,  which  is  the  absolute 
measure. 

E* 


'W^:":. 


54 


GEOMETRY. 


Scholium  2.  All  that  has  been  demonstrated  in  the  last  three 
propositions,  concerning  the  comparison  of  angles  with  arcs, 
holds  true  equally,  if  applied  to  the  comparison  of  sectors  with 
arcs ;  for  sectors  are  not  only  equal  when  their  angles  are  so, 
but  are  in  all  respects  proportional  to  their  angles ;  hence,  two 
sectors  ACB,  ACD,  taken  in  the  same  circle,  or  in  equal  circles^ 
are  to  each  other  as  the  arcs  AB,  AD,  the  bases  of  those  sectors. 
It  is  hence  evident  that  the  arcs  of  the  circle,  which  serve  as  a 
measure  of  the  different  angles,  are  proportional  to  the  different 
sectors,  in  the  same  circle,  or  in  equal  circles. 


PROPOSITION  XVIII.     THEOREM. 

An  inscribed  angle  is  measui^ed  by  half  the  arc  included  between 

its  sides. 


Let  BAD  be  an  inscribed  angle,  and  let 
us  first  suppose  that  the  centre  of  the  cir- 
cle lies  within  the  angle  BAD.  Draw  the 
diameter  AE,  and  the  radii  CB,  CD. 

The  angle  BCE,  being  exterior  to  the 
triangle  ABC,  is  equal  to  the  sum  of  the 
two  interior  angles  CAB,  ABC  (Book  I. 
Prop.  XXV.  Cor.  6.) :  but  the  triangle  BAC 
being  isosceles,  the  angle  CAB  is  equal  to 
ABC  ;  hence  the  angle  BCE  is  double  of  BAC.  Since  BCE  lies 
at  the  centre,  it  is  measured  by  the  arc  BE  ;  hence  BAC  will  be 
measured  by  the  half  of  BE.  For  a  like  reason,  the  angle  CAD 
will  be  measured  by  the  half  of  ED;  hence  BAC  +  CAD,  or  BAD 
will  be  measured  by  half  of  BE  +  ED,  or  of  BED. 

Suppose,  in  the  second  place,  that  the 
centre  C  lies  without  the  angle  BAD.  Then 
drawing  the  diameter  AE,  the  angle  BAE 
will  be  measured  by  the  half  of  BE  ;  the 
angle  DAE  by  the  half  of  DE  :  hence  their 
difference  BAD  will  be  measured  by  the 
half  of  BE  minus  the  half  of  ED,  or  by  the 
half  of  BD. 

Hence  every  inscribed  angle  is  measured 
bv  half  of  the  arc  included  between  its  sides. 


BOOK  III. 


55 


Cor.  1.  All  the  angles  BAG,  BDC, 
BEC,  inscribed  in  the  same  segment  are 
equal ;  because  they  are  all  measured  by 
the  half  of  the  same  arc  BOC. 


Cor.  2.  Every  angle  BAD,  inscribed  in  a 
semicircle  is  aright  angle  ;  because  it  is  mea- 
sured by  half  the  semicircumference  BOD, 
that  is,  by  the  fourth  part  of  the  whole  cir- 
cumference. 

Cor.  3.  Every  angle  BAG,  inscribed  in  a 
segment  greater  than  a  semicircle,  is  an  acute 
angle  ;  for  it  is  measured  by  half  of  the  arc 
BOC,  less  than  a  semicircumference. 

And    every  angle  BOG,  inscribed   in  a 
segment  less  than  a  semicircle,  is  an  obtuse 
angle  ;  for  it  is  measuf  ed  by  half  of  the  arc  ^ 
BAG,  greater  than  a  semicircumference. 


Cor.  4.  The  opposite  angles  A  and  G,  of 
aa  inscribed  quadrilateral  ABGD,  are  to- 
gether equal  to  tvv^o  right  angles  :  for  the  an- 
gle BAD  is  measured  by  half  the  arc  BGD, 
the  angle  BGD  is  measured  by  half  the  arc 
BAD  ;  hence  the  tvro  angles  BAD,  BGD,  ta- 
ken together,  are  measured  by  the  half  of  the 
circumference  ;  hence  their  sum  is  equal  to  two  right  angles. 


PROPOSITION  XIX.    THEOREM. 


The  angle  formed  by  two  chords,  which  intersect  each  other,  is 
measured  by  half  the  sum  of  the  arcs  included  between  its  sides 


56 


GEOMETRY. 


Let  AB,  CD, be  two  chords  intersecting 
each  other  at  E :  then  will  the  angle 
AEC,  or  DEB,  be  measured  by  half  of 
AC  +  DB. 

Draw  AF  parallel  to  DC :  then  will 
the  arc  DF  be  equal  to  AC  (Prop.  X.)  ; 
and  the  angle  FAB  equal  to  the  angle 
DEB  (Book  I.  Prop.  XX.  Cor.  3.).  But 
the  angle  FAB  is  measured  by  half  the 
arc  FDB  (Prop.  XVIII.);  therefore,  DEB 
is  measured  by  half  of  FDB  ;  that  is,  by  half  of  DB  +  DF,  or 
half  of  DB  + AC.  In  the  same  manner  it  might  be  proved  that 
the  angle  AED  is  measured  by  half  of  AFD  +  BC. 


PROPOSITION  XX.    THEOREM. 


The  angle  formed  hy  two  secants,  is  measured  by  half  the  diffe- 
i^ence  of  the  arcs  included  between  its  sides. 


Let  AB,  AC,  be  two  secants  :  then 
will  the  angle  BAC  be  measured  by 
half  the  difference  of  the  arcs  BEC 
and  DF. 

Draw  DE  parallel  to  AC  :  then  will 
the  arc  EC  be  equal  to  DF,  and  the 
angle  BDE  equal  to  the  angle  BAC. 
But  BDE  is  measured  by  half  the  arc 
BE ;  hence,  BAC  is  also  measured  by 
half  the  arc  BE  ;  that  is,  by  half  the 
difference  of  BEC  and  EC,  or  half  the 
difference  of  BEC  and  DF. 


PROPOSITION  XXI.    THEOREM. 


The  angle  formed  by  a  tangent  and  a  chord,  is  measured  by  half 
of  the  arc  included  between  its  sides. 


BOOK  III. 


57 


Let  BE  be  the  tangent,  and  AC  the  chord. 

From  A,  the  point  of  contact,  draw  the 
diameter  AD.  The  angle  BAD  is  a  right 
angle  (Prop.  IX.),  and  is  measured  by 
half  the  semicircumference  AMD ;  the 
angle  DAC  is  measured  by  the  half  of 
DC:  hence,  BAD  +  DAC,  or  PAC,  is 
measured  by  the  half  of  AMD  plus  the 
half  of  DC,  or  by  half  the  whole  arc 
AMDC. 

It  might  be  shown,  by  taking  the  difference  between  the  an- 
gles DAE,  DAC,  that  the  angle  CAE  is  measured  by  half  the 
arc  AC,  included  between  its  sides. 


-«.e@o«N- 


PROBLEMS  RELATING  TO  THE  FIRST  AND  THIRD  BOOKS 


PROBLEM  I. 
To  divide  a  given  straight  line  into  two  equal  parts,  '  * 


* 


><* 


Let  AB  be  the  given  straight  line. 

From  the  points  A  and  B  as  centres,  with 
a  radius  greater  than  the  half  of  AB,  describe 
two  arcs  cutting  each  other  in  D  ;  the  point 
D  will  be  equally  distant  from  A  and  B.  Find, 
in  like  manner,  above  or  beneath  the  line  AB,  - 
a  second  point  E,  equally  distant  from  the 
points  A  and  B ;  through  the  two  points  D 
and  E,  draw  the  line  DE :  it  will  bisect  the 
line  AB  in  C. 

For,  the  two  points  D  and  E,  being  each  equally  distant  from 
the  extremities  A  and  B,  must  both  lie  in  the  perpendicular 
raised  from  the  middle  of  AB  (Book  I.  Prop.  XVI.  Cor.).  But 
only  one  straight  line  can  pass  through  two  given  points  ;  hence 
the  line  DE  must  itself  be  that  perpendicular,  which  divides 
AB  into  two  equal  parts  at  the  point  C. 


X^ 


59 


GEOMETRY. 


PROBLEM  II. 

At^  given  pointy  in  a  given  straight  line,  to  erect  a  perpendicu- 
lar to  this  line. 


>N> 


■^ 


Let  A  be  the  given  point,  and  BC  the 
given  line. 

Take  the  points  B  and  C  at  equal  dis- 
tances from  A  ;  then  from  the  points  B  and 
C  as  centres,  vi^ith  a  radius  greater  than 
BA,  describe  two  arcs  intersecting  each 
other  in  D ;  draw  AD  :  it  will  be  the  perpendicular  required. 

For,  the  point  D,  being  equally  distant  from  B  and  C,  must 
be  in  the  perpendicular  raised  from  the  middle  of  BC  (Book  I. 
Prop.  XVI.) ;  and  since  two  points  determine  a  hne,  AD  is  that 
perpendicular. 

Scholium,  The  same  construction  serves  for  making  a  right 
angle  BAD,  at  a  given  point  A,  on  a  given  straight  line  BC. 


PROBLEM  III. 

From  a  given  point,  without  a  straight  line,  to  let  fall  a  perpen- 
dicular on  this  line. 


Let  Abe  the  point,  and  BD  the  straight 
line. 

From  the  point  A  as  a  centre,  and  with 
a  radius  sufficiently  great,  describe  an 
arc  cutting  the  line  BD  in  the  two  points 
B  and  D  ;  then  mark  a  point  E,  equally 
distant  from  the  points  B  and  D,  and 
draw  AE  :  it  will  be  the  perpendicular  required. 

For,  the  two  points  A  and  E  are  each  equally  distant  from 
the  points  B  and  D ;  hence  the  line  AE  is  a  perpendicular 
passing  through  the  middle  of  BD  (Book  I.  Prop.  XVL  Cor.). 


PROBLEM  IV. 


At  a  point  in  a  given  line,  to  make  an  angle  equal  to  a  given 

angle. 


BOOK  III.  59 

Let  A  be  the  given  point,  AB  the  given  line,  and  IKL  the 
given  angle. 

From  the  vertex  K,  as  a  cen-  t  o<^ 

tre,  with  any  radius,  describe  the  ^/^1\  .^-^vv 

arc  IL,  terminating  in  the  two-        ^^    \      ^^^    \ 

sides  of  the  angle.     From  the    '^^^ 1    .A  B~ 

point  A  as  a  centre,  with  a  dis- 
tance AB,  equal  to  Kl,  describe  the  indefinite  arc  BO  ;  then 
take  a  radius  equal  to  the  chord  LI,  with  which,  from  the  point 
B  as  a  centre,  describe  an  arc  cutting  the  indefinite  arc  BO,  in 
D ;  draw  KD ;  and  the  angle. DAB  will  be  equal  to  the  given 
angle  K. 

For,  the  two  arcs  BD,  LI,  have  equal  radii,  and  equal  chords ; 
hence  they  are  equal  (Prop.  IV.) ;  therefore  the  angles  BAD, 
IKL,  measured  by  them,  are  equal. 


PROBLEM  V. 
To  divide  a  given  arc,  or  a  given  angle,  into  two  equal  parts. 

First.  Let  it  be  required  to  divide  the 
arc  AEB  into  two  equal  parts.  From  the 
points  A  and  B,  as  centres,  with  the  same 
radius,  describe  two  arcs  cutting  each  other 
in  D  ;  through  the  point  D  and  the  centre 
C,  draw  CD :  it  will  bisect  the  arc  AB  in 
the"  point  E. 

For,  the  two  points  C  and  D  are  each 
equally  dptant  from  the  extremities  A  and  ^(^ 

B  of  the  Aord  AB  ;  hence  the  line  CD  bi- 
sects the^hord  at  right  angles   (Book  I.  Prop.  XVI.  Cor.)  ; 
hence,  it  bisects  the  arc    AB  in  the    point  E   (Prop.  VI.). 

Secondly.  Let  it  be  required  to  divide  the  angle  ACB  into 
two  equal  parts.  We  begin  by  describing,  from  the  vertex  C 
as  a  centre,  the  arc  AEB ;  w^iich  is  then  bisected  as  above.  It 
is  plain  that  the  line  CD  will  divide  the  angle  ACB  into  two 
equal  parts. 

Scholium.  By  the  same  construction,  each  of  the  halves 
AE,  EB,  may  be  divided  into  two  equal  parts  ;  and  thus,  by 
successive  subdivisions,  a  given  angle,  or  a  given  arc  may 
be  divided  into  four  equal  parts,  into  eight,  into  sixteen, 
and  so  on. 


ao 


GEOMETRY. 


PROBLEM  VI. 

Through  a  given  point,  to  draw  a  parallel  to  a  given  straight 

line. 

Let  A  be  the  given  point,  and  BC         |;^  -p^ 

the  given  hne.  JS 

From  the  point  A  as  a  centre,  with 
a  radius  greater  than  the  shortest  dis- 
tance from  A  to  BC,  describe  the  in-  Jl  i>I 
definite  arc  EO  ;  from  the  point  E  as  O 
a  centre,  with  the  same  radius,  describe  the  arc  AF  ;  make 
ED— AF,  and  draw  AD :  this  will  be  the  parallel  required. 

For,  drawing  AE,  the  alternate  angles  AEF,  EAD,  are  evi- 
dently equal ;  therefore,  the  lines  AD,  EF,  are  parallel  (Book  1. 
Prop.  XIX.  Cor.  1.). 


PROBLEM  VIL 

Two  angles  of  a  triangle  being  given,  to  find  the  third, 

Dr^wthe  indefinite  line  DEF; 
at  any  point  as  E,  make  the  an- 
gle DEC  equal  to  one  of  the 
given  angles,  and  the  angle 
CEH  equal  to  the  other  :  the 
remaining  angle  HEF  will  be 
the  third  angle  required ;  be- 
cause those  three  angles  are 
together  equal  to  two  right  angles  (Book  I.  Prop.  I. 
XXV). 


and 


PROBLEM  VIII. 


Two  sides  of  a  triangle,  and  the  angle  which  they  contain,  being 
given,  to  describe  the  triangle, 

Ltt  the  lines  B  and  C  be  equal  to 
the  given  sides,  and  A  the  given  an- 

Having  drawn  the  indefinite  line 
DE,  at  the  point  D,  make  the  angle        _ 
EDF  equal  to  the  given  angle  A ;    ^    ^ 
then  take  DG=B,  DH=C,  and  draw  GH  ;  DGH  will  be  the 
triangle  required  (Book  I.  Prop.  V.). 


BOOK  III. 


01 


PROBLEM  IX. 


A  side  and  two  angles  of  a  triangle  being  given,  to  describe  the 


triangle. 


The  two  angles  will  either  be  both  ad- 
jacent to  the  given  side,  or  the  one  adja- 
cent, and  the  other  opposite  :  in  the  lat- 
ter case,  find  the  third  angle  (Prob. 
VII.) ;  and  the  two  adjacent  angles  will 
thus  be  known  :  draw  the  straight  line 
DE  equal  to  the  given  side  :  at  the  point  D,  make  an  angle 
EDF  equal  to  one  of  the  adjacent  angles,  and  at  E,  an  angle 
DEG  equal  to  the  other ;  the  two  lines  DF,  EG,  will  cut  each 
other  in  H  ;  and  DEH  will  be  the  triangle  required  (Book  I. 
Prop.  VI.). 


PROBLEM  X. 


The  three  sides  of  a  triangle  being  given,  to  describe  the  triangle. 

Let  A,  B,  and  C,  be  the  sides. 

Draw  DE  equal  to  the  side  A : 
from  the  point  E  as  a  centre,  with 
a  radius  equal  to  the  second  side  B, 
describe  an  arc  ;  from  D  as  a  cen- 
tre, with  a  radius  equal  to  the  third 
side  C,  describe  another  arc  inter- 
secting the  former  in  F ;  draw  DF, 
EF  ;  and  DEF  will  be  the  triangle 
required  (Book  I.  Prop.  X.). 

Scholium.  If  one  of  the  sides  were  greater  than  the  sum  of 
the  other  two,  the  arcs  would  not  intersect  each  other  :  but  the 
solution  will  always  be  possible,  when  the  sum  of  two  sides,  any 
how  taken,  is  greater  than  the  third. 


F 


GEOMETRY. 


PROBLEM   XI. 


Two  sides  of  a  triangle^  and  the  angle  opposite  one  of  them,  being 
given,  to  describe  the  triangle. 

Let  A  and  B  be  the  given  sides,  and  C  the  given  angle. 
There  are  two  cases. 

First.  When  the  angle  C  is  a  right 
angle,  or  when  it  is  obtuse,  make 
the  angle  EDF=C;  take  DE=A  ; 
from  the  point  E  as  a  centre, 
with  a  radius  equal  to  the  given 
side  B,  describe  an  arc  cutting  DF 
in  F;  draw  EF :  then  DEF  will  be 
the  triangle  required. 

In  this  first  case,  the  side  B  must 
be  greater  than  A  ;  for  the  angle  C, 
being  a  right  angle,  or  an  obtuse  an- 
gle, is  the  greatest  angle  of  the  tri- 
angle, and  the  side  opposite  to  it  must,  therefore,  also  be  the 
greatest  (Book  I.  Prop.  XIIL). 


Ai- 


Secondly,  If  the  angle  C  is 
acute,  and  B  greater  than  A,  the 
^ame  construction  will  again  ap- 
ply, and  DEF  will  be  the  triangle 
required. 


But  if  the  angle  C  is  acute,  and 
the  side  B  less  than  A,  then  the 
arc  described  from  the  centre  E, 
with  the  radius  EF=B,  will  cut 
the  side  DF  in  two  points  F  and 
G,  lying  on  the  same  side  of  D : 
hence  there  will  be  two  triangles 
DEF,  DEG,  either  of  which  will 
satisfy  the  conditions  of  the  pro- 
blem. 


Bh 


SchoUum,  If  the  arc  described  with  E  as  a  centre,  should 
be  tangent  to  the  line  DG,  the  triangle  would  be  right  angled, 
and  there  would  be  but  one  solution.  The  problem  would  be 
impossible  in  all  cases,. if  the  side  B  were  less  than  the  perpen- 
dicular let  fall  from  E  on  the  line  DF, 


BOOK  III.  6B 


PROBLEM  XII. 

T/ie  adjacent  sides  of  a  parallelogram,  with  the  angle  which  they 
contain,  being  given,  to  describe  the  parallelogram. 

Let  A  and  B  be  the  given  sides,  and  C  the  given  angle. 

Drav^rthe  hne  DE=A;  at  i\vi 
point  D,  make  the  angle  EDF— 
C  ;  take  DF=B  ;  describe  two 
arcs,  the  one  from  F  as  a  cen- 
tre, with  a  radius  FG=DE,  the 


L 


other  from  E  as  a  centre,  with 

a  radius  EG=DF;  to  the  point  Ai — i 

G,  where  these  arcs  intersect  b  i 1 

each  other,  draw    FG,    EG ; 

DEGF  will  be  the  parallelogram  required. 

For,  the  opposite  sides  are  equal,  by  construction ;  hence  the 
figure  is  a  parallelogram  (Book  I.  Prop.  XXIX.)  :  and  it  is 
formed  with  the  given  sides  and  the  given  angle. 

Cor,  If  the  given  angle  is  a  right  angle,  the  figure  will  be 
a  rectangle  ;  if,  in  addition  to  this,  the  sides  are  equal,  it  will 
be  a  square. 


PROBLEM  Xin. 

To  find  the  centre  of  a  given  circle  or  arc. 

Take  three  points.  A,  B,  C,  any- 
where in  the  circumference,  or  the 
arc;  drawAB,BC,  or  suppose  them 
to  be  drawn ;  bisect  those  two  lines 
by  the  perpendiculars  DE,  FG : 
the  point  O,  where  these  perpen- 
diculars meet,  will  be  the  centre 
sought  (Prop.  VI.  Sch.). 

Scholium.  The  same  construc- 
tion sei-ves  for  making  a  circum- 
ference pass  through  three  given  points  A,  B,  C  ;  and  also  for 
describing  a  circumference,  in  which,  a  given  triangle  ABC 
shall  be  inscribed. 


CA 


GEOMETRY. 


PROBLEM  XIV. 
Through  a  given  pointy  to  draw  a  tangent  to  a  given  circle. 


If  the  given  point  A  lies  in  the  circum- 
ference, draw  the  radius  CA,  and  erect 
AD  perpendicular  to  it :  AD  will  be  the 
tangent  required  (Prop.  IX.). 


If  the  point  A  lies  without  the  circle, 
join  A  and  the  centre,  by  the  straight 
line  CA :  bisect  CA  in  O ;  from  O  as  a 
centre,  with  the  radius  OC,  describe  a 
circumference  intersecting  the  given  cir- 
cumference in  B ;  draw  AB  :  this  will  be 
the  tangent  required. 

For,  drawing  CB,  the  angle  CBA  be- 
ing inscribed  in  a  semicircle  is  a  right 
angle  (Prop.  XVIII.  Cor.  2.)  ;  therefore 
AB  is  a  perpendicular  at  the  extremity 
of  the  radius  CB ;  therefore  it  is  a  tan- 
gent. 

Scholium.  When  the  point  A  lies  without  the  circle,  there 
w^ill  evidently  be  always  two  equal  tangents  AB,  AD,  passing 
through  the  point  A :  they  are  equal,  because  the  right  angled 
triangles  CBA,  CDA,  have  the  hypothenuse  CA  common,  and 
the  side  CB  =  CD;  hence  they  are  equal  (Book  I.  Prop.  XVII.); 
hence  AD  is  equal  to  AB,  and  also  the  angle  CAD  to  CAB. 
And  as  there  can  be  but  one  line  bisecting  the  angle  BAC,  it 
follows,  that  the  line  which  bisects  the  angle  formed  by  two 
tangents,  must  pass  through  the  centre  of  the  circle. 


PROBLEM  XV. 

To  inscribe  a  circle  in  a.  given  triangle. 


Let  ABC  be  the  given  triangle. 

Bisect  the  angles  A  and  B,  by 
the  lines  AO  and  BO,  meeting  in 
the  point  O  ;  from  the  point  O, 
let  fall  the  perpendiculars  OD, 
OE,  OF,  on  the  three  sides  of  the 
triangle:  these  perpendiculars  will 
all  be  equal.     For,  by  construe- 


BOOK  III.  65 

lion,  we  have  the  angle  DAO=OAF,  the  right  angle  ADO= 
AFO  ;  hence  the  third  angle  AOD  is  equal  to  the  third  AOF 
(Book  I.  Prop.  XXV.  Cor.  2.).  Moreover,  the  side  AO  is  com- 
mon to  the  tvi^o  triangles  AOD,  AOF ;  and  the  angles  adjacent 
to  the  equal  side  are  equal :  hence  the  triangles  themselves  are 
equal  (Book  I.  Prop.  VI.) ;  and  DO  is  equal  to  OF.  In  the  same 
manner  it  may  be  shown  that  the  two  triangles  BOD,  BOE, 
are  equal ;  therefore  OD  is  equal  to  OE  ;  therefore  the  three 
perpendiculars  OD,  OE,  OF,  are  all  equal. 

Now,  if  from  the  point  O  as  a  centre,  with  the  radius  OD, 
a  circle  be  described,  this  circle  will  evidently  be  inscribed  in 
the  triangle  ABC  ;  for  the  side  AB,  being  perpendicular  to  the 
radius  at  its  extremity,  is  a  tangent ;  and  the  same  thing  is  true 
of  the  sides  BC,  AC, 

Scholium.  The  three  lines  which  bisect  the  angles  of  a  tri- 
angle meet  in  the  same  point. 


PROBLEM  XVI. 

On  a  given  straight  line  to  describe  a  segment  that  shall  contain 
a  given  angle ;  that  is  to  say,  a  segment  such,  that  all  the  an- 
gles inscribed  in  it,  shall  be  equal  to  the  given  angle. 

Let  AB  be  the  given  straight  line,  and  C  the  given  angle. 


Produce  AB  towards  D ;  at  the  point  B,  make  the  angle 
DBE=C;  draw  BO  perpendicular  to  BE,  and  GO  perpen- 
dicular to  AB,  through  the  middle  point  G ;  and  from  the  point 
O,  where  these  perpendiculars  meet,  as  a  centre,  with  a  dis- 
tance OB,  describe  a  circle :  the  required  segment  will  be 
AMB. 

For,  since  BF  is  a  perpendicular  at  the  extremity  of  the 
radius  OB,  it  is  a  tangent,  and  the  angle  ABF  is  measured  by 
half  the  arc  AKB  (Prop.  XXL).  Also,  the  angle  AMB,  being 
an  inscribed  angle,  is  measured  by  half  the  arc  AKB  :  hence 
we  have  AMB=ABF=:EBD=C :  hence  all  the  angles  in- 
scribed in  the  segment  AMB  are  equal  to  the  given  angle  C. 

F* 


66  GEOMETRY. 

Scholium.  If  the  given  angle  were  a  right  angle,  the  required 
segment  would  be  a  semicircle,  described  on  AB  as  a  diameter. 


PROBLEM  XVII. 

To  find  the  numerical  ratio  of  two  given  straight  lineSy  these  lines 
being  supposed  to  have  a  common  measure. 

Let  AB  and  CD  be  the  given  lines.  A    C 

From  the  greater  AB  cut  off  a  part  equal  to  the  less 
CD,  as  many  times  as  possible  ;  for  example,  twice, 
with  the  remainder  BE. 

From  the  line  CD,  cut  off  a  part  equal  to  the  re- 
mainder BE,  as  many  times  as  possible  ;  once,  for  ex- 
ample, with  the  remainder  DF. 

From  the  first  remainder  BE,  cut  off  a  part  equal  to 
the  second  DF,  as  many  times  as  possible  ;  once,  for 
example,  with  the  remainder  BG. 

From  the  second  remainder  DF,  cut  off  a  part  equal    \.q. 
to  BG  the  third,  as  many  times  as  possible. 

Continue  this  process,  till  a  remainder  occurs,  which    ^ 
is  contained  exactly  a  certain  number  of  times  in  the  preced- 
ing one. 

Then  this  last  remainder  will  be  the  common  measure  of  the 
proposed  lines ;  and  regarding  it  as  unity,  we  shall  easily  find 
the  values  of  the  preceding  remainders ;  and  at  last,  those  of 
the  two  proposed  lines,  and  hence  their  ratio  in  numbers. 

Suppose,  for  instance,  we  find  GB  to  be  contained  exactly 
twice  in  FD  ;  BG  w^ill  be  the  common  measure  of  the  two  pro- 
posed lines.  Put  BG=1  ;  we  shall  have  FD  =  2  :  but  EB  con- 
tains FD  once,  plus  GB ;  therefore  we  have  EB  =  3  :  CD  con- 
tains EB  once,  plus  FD ;  therefore  we  have  CD=:5  :  and, 
lastly,  AB  contains  CD  twice,  plus  EB  ;  therefore  we  have 
AB  =  13  ;  hence  the  ratio  of  the  hues  is  that  of  13  to  5.  If  the 
line  CD  were  taken  for  unity,  the  line  AB  would  be  ^^  ;  if  AB 
were  taken  for  unity,  CD  would  be  ^3. 

Scholium,  The  method  just  explained  is  the  same  as  that 
employed  in  arithmetic  to  find  the  common  divisor  of  two  num- 
bers ;  it  has  no  need,  therefore,  of  any  other  demonstration. 

How  far  soever  the  operation  be  continued,  it  is  possible 
that  no  remainder  may  ever  be  found,  which  shall  be  contained 
an  exact  number  of  times  in  the  preceding  one.  When  this 
happens,  the  two  lines  have  no  common  measure,  and  are  said 
to  be  int)ommensurable.  An  instance  of  this  will  be  seen  after- 


BOOK  III.  C7 

wards,  in  the  ratio  of  the  diagonal  to  the  side  of  the  squaro. 
In  those  cases,  therefore,  the  exact  ratio  in  numbers  cannot  be 
found  ;  but,  by  neglecting  the  last  remainder,  an  approximate 
ratio  will  be  obtained,  more  or  less  correct,  according  as  the 
operation  has  been  continued  a  greater  or  less  number  of  times. 


PROBLEM  XVIII. 

Two  angles  being  given,  to  find  their  common  measure,  if  they 
have  one,  and  by  means  of  it,  their  ratio  in  numbers. 

Let  A  and  B  be  the  given  an- 
gles. 

With  equal  radii  describe  the 
arcs  CD,  EF,  to  serve  as  mea- 
sures for  the  angles  :  proceed 
afterwards  in  the  comparison  of 
the  arcs  CD,  EF,  as  in  the  last 

problem,  since  an  arc  may  be  cut  off  from  an  arc  of  the  same 
radius,  as  a  straight  line  from  a  straight  line.  We  shall  thus 
arrive  at  the  common  measure  of  the  arcs  CD,  EF,  if  they  have 
one,  and  thereby  at  their  ratio  in  numbers.  This  ratio  will  be 
the  same  as  that  of  the  given  angles  (Prop.  XVII.) ;  and  if  DO 
is  the  common  measure  of  the  arcs,  DAO  will  be  that  of  the 
angles. 

Scholium.  According  to  this  method,  the  absolute  value  of 
an  angle  may  be  found  by  comparing  the  arc  which  measures 
it  to  the  whole  circumference.  If  the  arc  CD,  for  example,  is 
to  the  circumference,  as  3  is  to  25,  the  angle  A  will  be  /^  of  four 
right  angles,  or  if  of  one  right  angle. 

It  may  also  happen,  that  the  arcs  compared  have  no  com- 
mon measure  ;  in  which  case,  the  numerical  ratios  of  the  angles 
will  only  be  found  approximatively  with  more  or  less  correct- 
ness, according  as  the  operation  has  been  continued  a  greater 
or  less  number  of  times. 


68 


GEOMETRY. 


%■ 


BOOK  IV. 


OF  THE  PROPORTIONS  OF  FIGURES,  AND  THE  MEASUREMI>i>{T 
OF  AREAS. 


Definitions, 


1.  Similar  figures  are  those  which  have  the  angles  of  the  one 
equal  to  the  angles  of  the  other,  each  to  each,  and  the  sides 
about  the  equal  angles  proportional. 

2.  Any  two  sides,  or  any  two  angles,  which  have  like  po- 
sitions in  two  similar  figures,  are  called  homologous  sides  or 
angles. 

3.  In  two  different  circles,  similar  arcs,  sectors,  or  segments, 
are  those  which  correspond  to  equal  angles  at  the  centre. 

Thus,  if  the  angles  A  and  O  are  equal, 
the  arc  BC  will  be  similar  to  DE,  the 
sector  BAG  to  the  sector  DOE,  and  the 
segment  whose  chord  is  BC,  to  the  seg- 
ment whose  chord  is  DE. 

4.  The  base  of  any  rectilineal  figure,  is  the  side  on  which 
the  figure  is  supposed  to  stand. 

5.  The  altitude  of  a  triangle  is  the  per-  a. 
pendicular  let  fall  from  the  vertex  of  an 
angle  on  the  opposite  side,  taken  as  a 
base.     Thus,  AD  is  the  altitude  of  the 
triangle  BAG 


6.  The  altitude  of  a  parallelogram  is  the 
perpendicular  which  measures  the  distance 
between  two  opposite  sides  taken  as  bases. 
Thus,  EF  is  the  altitude  of  the  parallelo- 
gram DB. 

7.  The  altitude  of  a  trapezoid  is  the  per- 
pendicular drawn  between  its  two  parallel 
sides.  Thus,  EF  is  the  altitude  of  the  trape- 
zoid DB. 


DEC 


8.  The  area  and  surface  of  a  figure,  are  terms  very  nearly 
synonymous.  The  area  designates  more  particularly  the  super- 
ficial content  of  the  figure.    The  area  is  expressed  numeri- 


BOOK  IV.  69 

cally  by  the  number  of  times  wm'ch  the  figure  contains  some 
other  area,  that  is  assumed  for  its  measuring  unit. 

9.  Figures  have  equal  areas,  when  they  contain  the  same 
measuring  unit  an  equal  number  of  times. 

10.  Figures  which  have  equal  areas  are  called  equivalent. 
The  term  equal,  when  applied  to  figures,  designates  those  which 
are  equal  in  every  respect,  and  which  being  applied  to  each 
other  will  coincide  in  all  their  parts  (Ax.  13.)  :  the  term  equi- 
valent implies  an  equality  in  one  respect  only :  namely,  an 
equality  between  the  measures  of  figures. 

We  may  here  premise,  that  several  of  the  demonstrations 
are  grounded  on  some  of  the  simpler  operations  of  algebra, 
which  are  themselves  dependent  on  admitted  axioms.  Thus, 
if  we  have  A=B  +  C,  and  if  each  member  is  multiplied  by  the 
same  quantity  M,  we  may  infer  that  AxM=BxM  +  CxM; 
in  like  manner,  if  we  have,  A =6  + C,  and  D— E — C,  and  if  the 
equal  quantities  are  added  together,  then  expunging  the  +C 
and  — C,  which  destroy  each  other,  we  infer  that  A  +  D=B  + 
E,  and  so  of  others.  All  this  is  evident  enough  (Jf  itself;  but 
in  cases  of  difficulty,  it  will  be  useful  to  consult  some  agebrai- 
cal  treatise,  and  thus  to  combine  the  study  of  the  two  sciences. 


PROPOSITION  I.    THEOREM. 

Parallelograms  which  have  equal  bases  and  equal  altitudes^  are 
equivalent. 

Let  AB  be  the  common  base  of-j)      cp      EDPCE 
the    two   parallelograms  ABCD,    V       S/' 
ABEF  :  and  since  they  are  sup-      \ 
posed  to  have  the  same  altitude,        \ 
their  upper  bases  DC,  FE,  will  be        A       B  A        B 

both  situated  in  one  straight  line  parallel  to  AB. 

Now,  from  the  nature  of  parallelograms,  we  have  AD— BC, 
and  AF=BE;  for  the  same  reason,  we  have  DC=:AB,  and 
FE=AB  ;  hence  DC=rFE  :  hence,  if  DC  and  FE  be  taken 
away  from  the  same  line  DE,  the  remainders  CE  and  DF  will 
be  equal :  hence  it  follows  that  the  triangles  DAF,  CBE,  are 
mutually  eqilateral,  and  consequently  eqiial  (Book  I.  Prop.  X.). 

But  if  from  the  quadrilateral  ABED,  we  take  away  the  tri- 
angle ADF,  there  will  remain  the  parallelogram  ABEF ;  and 
if  from  the  same  quadrilateral  ABED,  we  take  away  the  equal 
triangle  CBE,  there  will  remain  the  parallelogram  ABCD. 


70  GEOMETRY. 

Hence  these  two  parallelograms  ABCD,  ABEF,  which  have 
the  same  base  and  altitude,  are  equivalent 

Cor,  Every  parallelogram  is  equivalent  to  the  rectangle 
which  has  the  same  base  and  the  same  altitude. 


PROPOSITION  II.    THEOREM. 

Evejy  triangle  is  half  the  parallelogram  which  has  the  same  base 
and  the  same  altitude. 

Let  ABCD  be  a  parallelo- 
gram, and  ABE  a  triangle, 
having  the  same  base  AB, 
and  the  same  altitude  :  then 
will  the  triangle  be  half  the 
parallelogram.  FA  B 

For,  since  the  triangle  and  the  parallelogram  have  the  same 
altitude,  the  vertex  E  of  the  triangle,  will  be  in  the  line  EC,  par- 
allel to  the  base  AB.  Produce  BA,  and  from  E  draw  EF 
parallel  to  AD.  The  triangle  FBE  is  half  the  parallelogram 
FC,  and  the  triangle  FAE  half  the  parallelogram  FD  (Book  I. 
Prop.  XXVIII.  Cor.). 

Now,  if  from  the  parallelogram  FC,  there  be  taken  the  par- 
allelogram FD,  there  will  remain  the  parallelogram  AC  :  and 
if  from  the  triangle  FBE,  which  is  half  the  first  parallelogram, 
there  be  taken  the  triangle  FAE,  half  the  second,  there  will  re- 
main the  triangle  ABE,  equal  to  half  the  parallelogram  AC. 

Cor  1 .  Hence  a  triangle  ABE  is  half  of  the  rectangle  ABGH, 
which  has  the  same  base  AB,  and  the  same  altitude  AH  :  for 
the  rectangle  ABGH  is  equivalent  to  the  parallelogram  ABCD 
(Prop.  I.  Cor.). 

Cor.  2.  All  triangles,  which  have  equal  bases  and  altitudes, 
are  equivalent,  being  halves  of  equivalent  parallelograms. 


PROPOSITION  III.    THEOREM. 

Two  rectangles  having  the  same  altitude^  are  to  each  other  as  their 

bases. 


BOOK  IV. 


71 


D 

1? 

( 

1 
J 

i     i 

1, 

= 

A 

E 

B 

Let  ABCD,  AEFD,  be  two  rectan- 
gles having  the  common  altitude  AD : 
they  are  to  each  other  as  their  bases 
AB,  AE. 

Suppose,  first,  that  the  bases  are 
commensurable,  and  are  to  each  other, 

for  example,  as  the  numbers  7  and  4.  If  AB  be  divided  into  7 
equal  parts,  AE  will  contain  4  of  those  parts :  at  each  point  of 
division  erect  a  perpendicular  to  the  base  ;  seven  partial  rect- 
angles will  thus  be  formed,  all  equal  to  each  other,  because  all 
have  the  same  base  and  altitude.  The  rectangle  ABCD  will 
:iontain  seven  partial  rectangles,  while  AEFD  will  contain  four: 
hence  the  rectangle  ABCD  is  to  AEFD  as  7  is  to  4,  or  as  AB 
is  to  AE.  The  same  reasoning  may  be  applied  to  any  other 
ratio  equally  with  that  of  7  to  4  :  hence,  whatever  be  that  ratio, 
if  its  terms  be  commensurable,  we  shall  have 

ABCD  :  AEFD  :  :  AB  :  AE. 

Suppose,  in  the  second  place,  that  the  bases 
AB,  AE,  are  incommensurable  :  it  is  to  be 
shown  that  we  shall  still  have 

ABCD  :  AEFD  :  :  AB  :  AE. 

For  if  not,  the  first  three  terms  continuing 
the  same,  the  fourth  must  be  greater  or  less 
than  AE.     Suppose  it  to  be  greater,  and  that  we  have 

ABCD  :  AEFD  :  :  AB  :  AO. 

Divide  the  line  AB  into  equal  parts,  each  less  than  EO. 
There  will  be  at  least  one  point  I  of  division  between  E  and 
O  :  from  this  point  draw  IK  perpendicular  to  AI :  the  bases 
AB,  AI,  will  be  commensurable,  and  thus,  from  what  is  proved 
above,  we  shall  have 

ABCD  :  AIKD  :  :  AB  :  AI. 

But  by  hypothesis  we  have 

ABCD  :  AEFD  :  :  AB  :  AO. 

In  these  two  proportions  the  antecedents  are  equal ;  hence 
the  consequents  are  proportional  (Book  II.  Prop.  IV.)  ;  and 
we  find 

AIKD  :  AEFD  :  :  AI  :  AO. 

But  AO  is  greater  than  AI ;  hence,  if  this  proportion  is  cor- 
rect, the  rectangle  AEFD  must  be  greater  than  AIKD :  on 
the  contrary,  however,  it  is  less  ;  hence  the  proportion  is  im- 
possible ;  therefore  ABCD  cannot  be  to  AEFD,  as  AB  is  to  a 
line  greater  than  AE. 


EIOB 


72  GEOMETRY. 

Exactly  in  the  same  manner,  it  may  be  shown  that  the  fourth 
term  of  the  proportion  cannot  be  less  than  AE  ;  therefore  it  is 
equal  to  AE. 

Hence,  whatever  be  the  ratio  of  the  bases,  two  rectangles 
ABCD,  AEFD,  of  the  same  altitude,  are  to  each  other  as  their 
bases  AB,  AE. 


PROPOSITION  IV.    THEOREM. 

Any  two  rectangles  are  to  each  other  as  the  products  of  their  bases 
multiplied  by  their  altitudes. 

Let  ABCD,  AEGF,  be  two  rectangles  ;  then  will  the  rect- 
angle, 

ABCD  :  AEGF  :  :  AB.AD  :  AF.AE. 

Having  placed  the  two  rectangles,  if  ^ "D 

so  that  the  angles  at  A  are  vertical, 
produce  the  sides  GE,  CD,  till  they 


B 


meet  in  H.  The  two  rectangles  ^ 
ABCD,  AEHD,  having  the  same  al- 
titude AD,  are  to  each  other  as  their  q 
bases  AB,  AE  :  in  like  manner  the 
two  rectangles  AEHD,  AEGF,  having  the  same  altitude  AE, 
are  to  each  other  as  their  bases  AD,  AF :  thus  we  have  the 
two  proportions, 

ABCD  :  AEHD  :  :  AB  :  AE, 
AEHD:  AEGF  :  :  AD  :  AF. 

Multiplying  the  corresponding  terms  of  these  proportions 
together,  and  observing  that  the  term  AEHD  may  be  omit- 
ted, since  it  is  a  multiplier  of  both  the  antecedent  and  the  con- 
sequent, we  shall  have 

ABCD  :  AEGF  :  :  ABxAD  :  AExAF. 

Scholium.  Hence  the  product  of  the  base  by  the  altitude  may 
be  assumed  as  the  measure  of  a  rectangle,  provided  we  under- 
stand by  this  product,  the  product  of  two  numbers,  one  of 
which  is  the  number  of  linear  units  contained  in  the  base,  the 
other  the  number  of  linear  units  contained  in  the  altitude.  This 
product  will  give  the  number  of  superficial  units  in  the  surface  ; 
because,  for  one  unit  in  height,  there  are  as  many  superficial 
units  as  there  are  linear  units  in  the  base ;  for  two  units  in 
height  twice  as  many ;  for  three  units  in  height,  three  times  as 
many,  &c. 

Still  this  measure  is  not  absolute,  but  relative  :  it  supposes 


BOOK  IV 


7a 


that  the  area  of  any  other  rectangle  is  computed  m  a  similar 
manner,  by  measuring  its  sides  with  the  same  linear  unit ;  a 
second  product  is  thus  obtained,  and  the  ratio  of  the  two  pro- 
ducts is  the  same  as  that  of  the  rectangles,  agreeably  to  the 
proposition  just  demonstrated. 

For  example,  if  the  base  of  the  rectangle  A  contains  three 
units,  and  its  altitude  ten,  that  rectangle  will  be  represented 
by  the  number  3x  10,  or  30,  a  number  which  signifies  nothing 
while  thus  isolated ;  but  if  there  is  a  second  rectangle  B,  the 
base  of  which  contains  twelve  units,  and  the  altitude  seven,  this 
second  rectangle  will  be  represented  by  the  number  12x7  = 
84  ;  and  we  shall  hence  be  entitled  to  Conclude  that  the  two 
rectangles  are  to  each  other  as  30  is  to  84  ;  and  therefore,  if 
the  rectangle  A  were  to  be  assumed  as  the  unit  of  measurement 
in  surfaces,  the  rectangle  B  would  then  have  |4  for  its  absolute 
measure,  in  other  words,  it  would  be  equal  to  ^^  of  a  super- 
ficial unit. 

It  is  more  common  and  more 
simple,  to  assume  the  square  as 
the  unit  of  surface  ;  and  to  se- 
lect that  square,  whose  side  is 
the  unit  of  length.  In  this  case 
the  measurement  which  we  have 

regarded  merely  as  relative,  becomes  absolute  :  the  number  30, 
for  instance,  by  which  the  rectangle  A  was  measured,  now 
represents  30  superficial  units,  or  30  of  those  squares,  which 
have  each  of  their  sides  equal  to  unity,  as  the  diagram  exhibits. 

In  geometry  the  product  of  two  lines  frequently  means  the 
same  thing  as  their  rectangle,  and  this  expression  has  passed 
into  arithmetic,  where  it  serves  to  designate  the  product  of  two 
unequal  numbers,  the  expression  square  being  employed  to 
designate  the  product  of  a  number  multiplied  by  itself. 

The  arithmetical  squares  of  1,  2,  3, 
&c.  are  1,  4,  9,  &c.  So  likewise,  the 
geometrical  square  constructed  on  a 
double  line  is  evidently  four  times 
greater  than  the  square  on  a  single  one ; 
on  a  triple  fine  it  is  nine  times  great- 
er,  i&c. 


A 


>^   Of 


74  GEOMETRY. 


PROPOSITION  V.    THEOREM. 

The  area  of  any  parallelogram  is  equal  to  the  product  of  its  base 
by  its  altitude. 

For,  the  parallelogram  ABCD  is  equivalent  3?  j) 
to  the  rectangle  ABEF,  which  has  the  same 
base  AB,  and  the  same  altitude  BE  (Prop.  I. 
Cor.) :  but  this  rectangle  is  measured  by  AB 
xBE  (Prop.  IV.  Sch.);  therefore,  ABxBE 
is  equal  to  the  area  of  the  parallelogram  ABCD. 

Cor.  Parallelograms  of  the  same  base  are  to  each  other  as 
their  altitudes ;  and  parallelograms  of  the  same  altitude  are  to 
each  other  as  their  bases :  for,  let  B  be  the  common  base,  and 
C  and  D  the  altitudes  of  two  parallelograms : 

then,        BxC:BxD::C:D,  (Book  II.  Prop.  VII.) 

And  if  A  and  B  be  the  bases,  and  C  the  common  altitude, 
we  shall  have 

AxC  :  BxC  :  :  A  :  B. 

And  parallelograms,  generally,  are  to  each  other  as  the  pro- 
ducts of  their  bases  and  altitudes. 


PROPOSITION  VI.    THEOREM. 

The  area  of  a  triangle  is  equal  to  the  product  of  its  base  by  half 
its  altitude. 

For,  the  triangle  ABC  is  half  of  the  par- 
allelogram ABCE,  which  has  the  same  base 
BC,  and  the  same  altitude  AD  (Prop.  II.) ; 
but  the  area  of  the  parallelogram  is  equal  to 
BC  X  AD  (Prop.  V.) ;  hence  that  of  the  trian- 
gle must  be  iBC  x  AD,  or  BC  x  ^AD. 

Cor.  Two  triangles  of  the  same  altitude  are  to  each  other  as 
their  bases,  and  two  triangles  of  the  same  base  are  to  each 
other  as  their  altitudes.  And  triangles  generally,  are  to  each 
other,  as  the  products  of  their  bases  and  altitudes. 


BOOK  iV.  76 


PROPOSITION  VII.  '  THEOREM. 

The  area  of  a  trapezoid  is  equal  to  its  altitude  multiplied  by  the 
half  sum  of  its  parallel  bases. 

Let  ABCD  be  a  trapezoid,  EF  its  alti- 
tude, AB  and  CD  its  parallel  bases ;  then 
will  its  area  be  equal  to  EFx  i(AB  +  CD). 

Through  I,  the  middle  point  of  the  side 
BC,  draw  KL  parallel  to  the  opposite  side 
AD  ;  and  produce  DC  till  it  meets  KL.        ^     ^ 

In  the  triangles  IBL,  ICK,  we  have  the  side  IB=IC,  by 
construction;  the  angle  LIB  =  CIK;  and  since  CK  and  BL 
are  parallel,  the  angle  IBL=ICK  (BookL  Prop.  XX.  Cor.  2.); 
hence  the  triangles  are  equal  (Book  L  Prop.  VL) ;  therefore, 
the  trapezoid  ABCD  is  equivalent  to  the  parallelogram  ADKL, 
and  is  measured  by  EF  x  AL. 

But  we  have  AL=DK ;  and  since  the  triangles  IBL  and 
KCI  are  equal,  the  side  BL=CK:  hence,  AB  + CD- AL + 
DK=2AL ;  hence  AL  is  the  half  sum  of  the  bases  AB,  CD ; 
hence  the  area  of  the  trapezoid  ABCD,  is  equal  to  the  altitude 
EF  multiplied  by  the  half  sum  of  the  bases  AB,  CD,  a  result 

which  is  expressed  thus:  ABCD=EFx^^^— . 

Scholium.  If  through  I,  the  middle  point  of  BC,  the  line  IH 
be  drawn  parallel  to  the  base  AB,  the  point  H  will  also  be  the 
middle  of  AD.  For,  since  the  figure  AHIL  is  a  parallelogram, 
as  also  DHIK,  their  opposite  sides  being  parallel,  we  have 
AH=IL,  and  DII=IK;  but  since  the  triangles  BIL,  CIK,  are 
equal,  we  already  have  IL=IK;  therefore,  AH  =  DH. 

It    may  be   observed,  that  the    line    HI=AL   is  equal  to 

— ;  hence  the  area  of  the  trapezoid  may  also  be  ex- 

pressed  by  EF  x  HI :  it  is  therefore  equal  to  the  altitude  of  the 
trapezoid  multiplied  by  the  line  which  connects  the  middle 
points  of  its  inclined  sides. 


76  GEOMETRY. 


PROPOSITION  VIII.    THEOREM. 

If  aline  is  divided  into  two  parts,  the  square  described  on  the 
whole  line  is  equivalent  to  the  sum  of  the  squares  described  on 
the  parts,  together  with  twice  the  rectangle  contained  by  the 
parts.  I 

Let  AC  be  the  line,  and  B  the  point  of  division  ;  then,  is 
AC2  or  (AB  +  BC)2=ABHBC2+2ABxBC. 

Construct   the  square  ACDE ;  take  AF=    S  H     JO 
AB ;  draw  FG  parallel  to  AC,  and  BH  par- 
allel to  AE.                                                            ^ fi —  O 

The  square  ACDE  is  made  up  of  four  parts  ; 
the  first  ABIF  is  the  square  described  on  AB, 
since  we  made  AF=AB :  the  second  IDGH  is    K  B    C 

the  square  described  on  IG,  or  BC  ;  for  since  we  have  AC  = 
AE  and  AB=AF,  the  difference,  AC — AB  must  be  equal  to 
the  difference  AE — AF,  which  gives  BC=EF  ;  but  IG  is  equal 
to  BC,  and  DG  to  EF,  since  the  lines  are  parallel ;  therefore 
IGDH  is  equal  to  a  square  described  on  BC.  And  those  two 
squares  being  taken  away  from  the  whole  square,  there  re- 
mains the  two  rectangles  BCGI,  EFIH,  each  of  which  is  mea- 
sured by  AB  X  BC  ;  hence  the  large  square  is  equivalent  to  the 
two  small  squares,  together  with  the  two  rectangles. 

Cor.  If  the  line  AC  were  divided  into  two  equal  parts,  the 
two  rectangles  EI,  IC,  would  become  squares,  and  the  square 
described  on  the  whole  line  would  be  equivalent  to  four  times 
the  square  described  on  half  the  line. 

Scholium.  This  property  is  equivalent  to  the  property  de- 
monstrated in  algebra,  in  obtaining  the  square  of  a  binominal ; 
which  is  expressed  thus  : 

(a-Vbf=a^-V2abVh\ 


PROPOSITION  IX.    THEOREM. 

The  square  described  on  the  difference  oftwe  lines,  is  equivalent 
to  the  sum  of  the  squares  described  on  the  Jhe^,  r^inus  twice 
the  rectangle  contained  by  the  lines. 


BOOK  IV. 


77 


B. 


Let  AB  and  BC  be  two  lines,  AC  their  difference  ;  then  is 
AC^,  or  (AE— BC)2=AB2+BC2— 2ABxBC. 

Describe  the  square  ABIF  ;  take  AE    Xi      ]? G-      I 

= AC  ;  draw  CG  parallel  to  to  BI,  HK 
parallel  to  AB,  and  complete  the  square 
£FLK. 

The  two  rectangles  CBIG,  GLKD, 
arc  each  measured  by  AB  x  BC  ;  take 

them    away  from    the    whole    figure  ^ 

ABILKEA,  which   is    equivalent   to  ^      ^ 

AB^-f  BC^  and  there  will  evidently  remain  the  square  ACDE; 
hence  the  theorem  is  true. 

Scholium,    This  proposition  is  equivalent  to  the  algebraicai 
formula,  (a—by=ar-~-2ab+b\ 


G    I 


PROPOSITION  X.    THEOREM. 

The  rectangle  contained  by  the  sum  and  the  difference  of  two 
lines,  is  equivalent  to  the  difference  of  the  squares  of  those 
lines. 

Let  AB,  BC,  be  two  lines ;  then,  will 

(AB+BC)  X  (AB— BC)=AB2— BC2. 

On  AB  and  AC,  describe  the  squares    n 
ABIF,  ACDE ;  produce  AB  till  the  pro- 
duced part  BK  is  equal  to  BC  ;  and  p, 
complete  the  rectangle  AKLE. 

The  base  AK  of  the  rectangle  EK, 
is  the  sum  of  the  two  lines  AB,  BC  ;  its 
altitude  AE  is  the  difference  of  the 
same  lines ;  therefore  the  rectangle  j 
AKLE  is  equal  to  (AB  +  BC)  x  (AB— 
BC).  But  this  rectangle  is  composed  of  the  two  parts  ABHE 
+  BHLK  ;  and  the  part  BHLK  is  equal  to  the  rectangle  EDGP, 
because  BH  is  equal  to  DE,  and  BK  to  EF  ;  hence  AKLE  is 
equal  to  ABHE  +  EDGF.  These  two  parts  make  up  the  square 
ABIF  minus  the  square  DHIG,  which  latter  is  equal  to  a 
square  described  on  BC  :  hence  wo  have 

(AB+BC)  X  (AB— BC)=AB2— BC2 

Scholium.     This  proposition  is  equivalent  to  the  algebraical 
formula,  (a+6)  x  (a — b)=a^ — b^. 


H 

D 

C     B     K 


G* 


78 


GEOMETRY. 


PROPOSITION  XI.    THEOREM 


The  square  described  on  the  hypothenuse  of  a  right  angled  tr>  - 
angle  is  equivalent  to  the  sum  of  the  squares  described  on  th4 
other  two  sides. 


Let  the  triangle  ABC  be  right 
angled  at  A.  Having  described 
squares  on  the  three  sides,  let 
fall  from  A,  on  the  hypothenuse, 
the  perpendicular  AD,  which 
produce  to  E;  and  draw  the 
diagonals  AF,  CH. 

The  angle  ABF  is  made  up 
of  the  angle  ABC,  together  with 
the  right  angle  CBF  ;  the  angle 
CBH  is  made  up  of  the  same 
angle  ABC,  together  with  the 
right  angle  ABH  ;   hence  the  -  ^  e 

angle  ABF  is  equal  to  HBC.  But  we  have  AB 
sides  of  the  same  square  ;  and  BF==BC,  for  the  same  reason 
therefore  the  triangles  ABF,  HBC,  have  two  sides  and  the  in- 
cluded angle  in  each  equal ;  therefore  they  are  themselves 
equal  (Book  I.  Prop.  V.). 

The  triangle  ABF  is  half  of  the  rectangle  BE,  because  they 
have  the  same  base  BF,  and  the  same  altitude  BD  (Prop.  II. 
Cor.  1.).  The  triangle  HBC  is  in  like  manner  half  of  the 
square  AH :  for  the  angles  BAC,  BAL,  being  both  right  angles, 
AC  and  AL  form  one  and  the  same  straight  line  parallel  to 
HB  (Book  I.  Prop.  HI.) ;  and  consequently  the  triangle  HBC, 
and  the  square  AH,  which  have  the  common  base  BH,  have 
also  the  common  altitude  AB ;  hence  the  triangle  is  half  of  the 
square. 

The  triangle  ABF  has  already  been  proved  equal  to  the  tri- 


BH,  being 


angle  HBC ; 


hence  the  rectangle  BDEF,  which  is  double  of 


the  triangle  ABF,  must  be  equivalent  to  the  square  AH,  which 
is  double  of  the  triangle  HBC.  In  the  same  manner  it  may  be 
proved,  that  the  rectangle^  CDEG  is  equivalent  to  the  square 
AI.  But  the  two  rectangles  BDEF,  CDEG,  taken  together, 
make  up  the  square  BCGF  :  therefore  the  square  BCGF,  de- 
scribed  on  the  hypothenuse,  is  equivalent  to  the  sum  of  tht 
squares  ABHL,  ACIK,  described  on  the  two  other  sides  ;  is 
other  words,  BC^^ABHAG^. 


BOOK  IV. 


79 


Co7\  1.  Hence  the  square  of  one  of  the  sides  of  a  right  an- 
gled triangle  is  equivalent  to  the  square  of  the  hypothenuse 
diminished  by  the  square  of  the  other  side  ;  which  is  Lius  ex- 
pressed :  AB2=BC2— AC2. 

Cor.  2.    It  has  just  been  shown  that  the  square  AH  is  equi- 
valent to  the  rectangle  BDEF ;  but  by  reason  of  the  common 
altitude  BF.  the  square  BCGF  is  to  the  rectangle  BDEF  as  the 
base  BC  is  to  the  base  BD ;  therefore  we  have 
BC2  :  AB2  :  :  BC  :  BD. 

Hence  the  square  of  the  hypothenuse  is  to  the  square  of  one  of 
the  sides  about  the  right  angle,  as  the  hypothenuse  is  to  the  seg- 
ment adjacent  to  that  side.  The  word  segment  here  denotes 
that  part  of  the  hypothenuse,  which  is  cut  off  by  the  perpen- 
dicular let  fall  from  the  right  angle  :  thus  BD  is  the  segment 
adjacent  to  the  side  AB  ;  and  DC  is  the  segment  adjacent  to 
the  side  AC.     We  might  have,  in  like  manner, 

BC2  :  AC2  :  :  BC  :  CD. 

Cor.  3.  The  rectangles  BDEF,  DCGE,  having  likewise  the 
same  altitude,  are  to  each  other  as  their  bases  BD,  CD.  But 
these  rectangles  are  equivalent  to  the  squares  AH,  AI ;  there- 
fore we  have  AB^  :  AC^  :  :  BD  :  DC. 

Hence  the  squares  of  the  two  sides  containing  the  right  angle, 
are  to  each  other  as  the  segments  of  the  hypothenuse  which  lie 
adjacent  to  those  sides.  ' 

Cor.  4.  Let  ABCD  be  a  square,  and  AC  its 
diagonal :  the  triangle  ABC  being  right  an- 
gled and  isosceles,  we  shall  have  AC^=AB^+ 
BC^=2AB^:  hence  the  square  described  on  the 
diagonal  AC,  is  double  of  the  square  described 
on  the  side  AB. 

This  property  may  be  exhibited  more  plainly, 
by  drawing  parallels  to  BD,  through  the  points  A  and  C,  and 
parallels  to  AC,  through  the  points  B  and  D.  A  new  square 
EFGH  will  thus  be  formed,  equal  to  the  square  of  AC.  Now 
EFGH  evidently  contains  eight  triangles  each  equal  to  ABE  ; 
and  ABCD  contains  four  such  triangles :  hence  EFGH  is 
double  of  ABCD. 

Since  we  have  AC^  :  AB^  :  :  2  :  1  ;  by  extractmg  the 
square  roots,  we  shall  have  AC  :  AB  :  :  \/2  :  1  ;  hence,  the 
diagonal  of  a  square  is  incommensurable  with  its  side  ;  a  pro- 
perty which  will  be  explained  more  fully  in  another  place. 


80 


GEOMETRY. 


PROPOSITION  XII.    THEOREM. 


In  every  triangle,  the  square  of  a  side  opposite  an  acute  angle  fa 
less  than  the  sum  of  the  squares  of  the  other  two  sides,  by  twice 
the  rectangle  contained  by  the  base  and  the  distance  from  the 
acute  angle  to  the  foot  of  the  perpendicular  let  fall  from  the 
opposite  angle  on  the  base,  or  on  the  base  produced. 

Let  ABC  be  a  triangle,  and  AD  perpendicular  to  the  base 
CB ;  then  will  AB^^  AC2+ BC^— 2BC  x  CD. 

There  are  two  cases. 

First.  When  the  perpendicular  falls  within 
the  triangle  ABC,  we  have  BDrrrBC— CD, 
and  consequently  BD2=BC2+CD^— 2BC 
xCD  (Prop.  IX.).  Adding  AD^  to  each, 
and  observing  that  the  right  angled  trian- 
gles ABD,  ADC,  give  ADHBD^^  AB^,  and 
ADHCD2=AC2,  we  have  AB^-BC^^-  _ 
AC^— 2BCxCD.  ^ 

Secondly.  When  the  perpendicular  AD 
falls  without  the  triangle  ABC,  we  have  BD 
=  CD— BC  ;  and  consequently  BD^^CD^-j- 
BC2— 2CD  X  BC  (Prop.  IX.).  Adding  AD^ 
to  both,  we  find,  as  before,  AB^^BCHAC^ 
— 2BCxCD. 


PROPOSITION  XIII.    THEOREM. 


In  every  obtuse  angled  triangle,  the  square  of  the  side  opposite  the 
obtuse  angle  is  greater  than  the  sum  of  the  squares  of  the  other 
two  sides  by  twice  the  rectangle  contained  by  the  base  and  the 
distance  from  the  obtuse  angle  to  the  foot  of  the  peipendicular 
let  fall  from  the  opposite  angle  on  the  base  produced. 

Let  ACB  be  a  triangle,  C  the  obtuse  angle,  and  AD  perpen- 
dicular to  BC  produced  ;  then  will  AB2=AC2+BC-+2BCx 
CD. 

The  perpendicular  cannot  fall  within  the 
triangle  ;  for,  if  it  fell  at  any  point  such  as 
E,  there  would  be  in  the  triangle  ACE,  the 
right  angle  E,  and  the  obtuse  angle  C,  which 
is  impossible  (Book  L  Prop.  XXV.  Cor.  3.) : 


BOOK  IV.  81 

hence  the  perpendicular  falls  without ;  and  we  have  BD=BC 
+  CD.  From  this  there  results  BD^^BCHCDHSBC  x  CD 
(Prop.  VIII.).  Adding  AD^  to  both,  and  reducing  the  sums  as 
in  the  last  theorem,  we  find  AB^^BCH  AC-H2BC  x  CD. 

Scholium.  The  right  angled  triangle  is  the  only  one  in  which 
the  squares  described  on  the  two  sides  are  together  equivalent 
to  the  square  described  on  the  third ;  for  if  the  angle  contained 
by  the  two  sides  is  acute,  the  sum  of  their  squares  will  be 
greater  than  the  square  of  the  opposite  side ;  if  obtuse,  it  will 
be  less. 

PROPOSITION  XIV.    THEOREM. 

In  any  triangle,  if  a  straight  line  he  drawn  from  the  vertex  to  the 
middle  of  the  base,  twice  the  square  of  this  line,  together  with 
twice  the  square  of  half  the  base,  is  equivalent  to  the  sum  of  the 
squares  of  the  other  two\sides  of  the  triangle. 

Let  ABC  be  any  triangle,  and  AE  a  line  drawn  to  the  mid- 
dle of  the  base  BC  ;  then  will 

2AEH2BE2=AB2+AC2. 
On  BC,  let  fa'l  the  perpendicular  ADr  A 

Then,  by  Prop.  XII. 

AC^riAEHEC^— SEC  x  ED. 
And  by  Prop.  XI 11. 

AB2-AEHEB2+2EB  x  ED.  j^^ ^-^ 

Hence,  by  adding,  and  observing  that  EB  and  EC  are  equal, 
we  have 

AB2 + AC2r=2  AE2  +  2EB2. 

Cor.  Hence,  in  every  parallelogram  the  squares  of  the  sides 
are  together  equivalent  to  the  squares  of  the  diagonals. 

For  the  diagonals  AC,  BD,  bisect  each  q  q 

other  (Book  I.  Prop.  XXXI.) ;  consequently 
the  triangle  ABC  gives  \  ""^^ 

AB^-f  BC2z=2AEH  2BE2. 

The  triangle  ADC  gives,  in  like  manner. 

AD2+  DC2-2AE2  ^  2DE2. 
Adding  the  corresponding  members  together,  and  observing  that 
BE  and  DE  are  equal,  we  shall  have  ^ 

AB2+AD2fDC2-fBC2n4AE2+4DE2.  >f> 

But  4AE2  is  the  square  of  2AE,  or  of  AC  ;  ^DW  is  the  square 
of  BD  (Prop.  VIII.  Cor.) :  hence  the  squares  of  the  sides  are 
together  equivalent  to  the  squares  of  the  diagonals. 


82 


GEOMETRY. 


PROPOSITION  XV.    THEOREM. 


If  a  line  he  drawn  parallel  to  the  base  of  a  triangle^  it  will  divide 
tfie  other  sides  proportionally, 

Eet  ABC  be  a  triangle,  and  DE  a  straight  line  drawn  par 
allel  to  the  base  BC  ;  then  will 

AD  :  DB  :  :  AE  :  EC. 

Draw  BE  and  DC.  The  two  triangles  BDE, 
DEC  having  the  same  base  DE,  and  the  same 
altitude,  since  both  their  vertices  lie  in  a  line 
parallel  to  the  base,  are  equivalent  (Prop.  II. 
Cor.  2.). 

'  The  triangles  ADE,  BDE,  whose  common 
vertex  is  E,  have  the  same  altitude,  and  are  to 
each  other  as  their  bases  (Prop.  VI.  Cor.)  ; 
hence  we  have 

ADE  ;  BDE  :  :  AD  :  DB. 

The  triangles  ADE,  DEC,  whose  common  vertex  is  D,  have 
also  the  same  altitude,  and  are  to  each  other  as  their  bases  ; 
hence 

ADE  :  DEC  :  :  AE  :  EC. 

But  the  triangles  BDE,  DEC,  are  equivalent ;  and  therefore, 
we  have  (Book  II.  Prop.  IV.  Cor.) 

AD  :  DB  :  :  AE  :  EC. 

Cor.  1.  Hence,  by  composition,  we  have  AD  +  DB  :  AD  :  : 
AE  +  EC  :  AE,  or  AB  :  AD  :  :  AC  :  AE  ;  and  also  AB  : 
BD  :  :  AC  :  CE. 

Cor.  2.  If  between  two  straight  lines  AB,  CD,  any  number 
of  parallels  AC,  EF,  GH,  BD,  &lc.  be  drawn,  those  straight 
lines  will  be  cut  proportionally,  and  we  shall  have  AE  :  CF  '> 
EG  :  FH  :  GB  :  HD. 

For,  let  O  be  the  point  where  AB  and 
CD  meet.  In  the  triangle  OEF,  the  line 
AC  being  drawn  parallel  to  the  base  EF, 
we  shall  have  OE  :  AE  :  :  OF  :  CF,  or 
OE  :  OF  :  :  AE  :  CF.  In  the  triangle 
OGH,  we  shall  likewise  have  OE  :  EG 
:  :  OF  :  FH,orOE  :  OF  :  :  EG  :  FH. 
And  by  reason  of  the  common  ratio  OE  : 
OF,  those  two  proportions  give  AE  :  CF 
:  :  EG  :  FH.  It  may  be  proved  in  the 
same  manner,  that  EG  ;  FH  :  :  GB  :  HD,  and  so  on  ;  hence 
the  lines  AB,  CD,  are  cut  proportionally  by  the  parallels  AC, 
EF,  GH,  i&c. 


1 


BOOK  IV. 


8d 


PROPOSITION  XVI.    THEOREM. 


Conversely,  if  two  sides  of  a  triangle  are  cut  proportionally  hy  a 
straight  line,  this  straight  line  will  be  parallel  to  the  third  side. 


In  the  triangle  ABC,  let  the  line  DE  be  drawn,  making 
AD  :  DB  :  :  AE  :  EC  :  then  will  DE  be  parallel  to  BC. 

For,  if  DE  is  not  parallel  to  BC,  draw  DO  paral- 
lel to  it.  Then,  by  the  preceding  theorem,  we  shall 
have  AD  :  DB  :  :  AO  :  OC.  But  by  hypothe- 
sis, we  have  AD  :  DB  :  :  AE  :  EC :  hence  we 
must  have  AO  :  OC  :  :  AE  :  EC,orAO  ;  AE 
:  :  OC  :  EC  ;  an  impossible  result,  since  AO,  the 
one  antecedent,  is  less  than  its  consequent  AE, 
and  OC,  the  other  antecedent,  is  greater  than  its 
consequent  EC.  Hence  the  parallel  to  BC,  drawn  from  the 
point  I),  cannot  differ  from  DE ;  hence  DE  is  that  parallel. 

Scholium.  The  same  conclusion  would  be  true,  if  the  pro- 
portion AB  :  AD  :  :  AC  :  AE  were  the  proposed  one.  For 
this  proportion  would  give  AB — AD  :  AD  :  :  AC — AE  : 
AE,  or  BD  :  AD  :  :  CE  :  AE. 


PROPOSITION  XVII.    THEOREM. 

The  line  which  bisects  the  vertical  angle  of  a  triangle,  divides  the 
base  into  two  segments,  which  are  proportional  to  the  adjacent 
sides. 


bisecting 


the  angle 


AC. 


E 


In  the  triangle  ACB,  let  AD  be  drawn 
CAB ;  then  will 

BD  :  CD  :  :  AB  : 

Through  the  point  C,  draw  CE 
parallel  to  AD  till  it  meets  BA 
produced. 

In  the  triangle  BCE,  the  line  AD 
is  parallel  to  the  base  CE ;  hence 
we  have  the  proportion  (Prop. 
XV.), 

BD  :  DC  :  :  AB  :  AE. 

But  the  triangle  ACE  is  isos- 
celes :  for,  since  AD,  CE  are  parallel,  we  have  the  angle  ACE 
=DAC,  and  the  angle  AEC=BAD  (Book  I.  Prop.  XX.  Cor. 
2  &  3.) ;  but,  by  hypothesis,  DACinBAD ;  hence  the  an- 
gle ACE— AEC,  and  consequently  AErrAC  (Book  I.  Prop. 
XII.).  In  place  of  AE  in  the  above  proportion,  substitute  AC, 
and  we  shall  have  BD  :  DC  :  ;  AB  :  AC. 


84  GEOMETRY. 


PROPOSITION  XVIII.     THEOREM. 

Two  equiangular  triangles  have  their  homologous  sides  propor- 
tional, and  are  similar. 

Let  ABC,  CDE  be  two  triangles  which  E 

have  their  angles  equal  each  to  each,  /\ 

namely,  BAC=CDE,  ABCz=:DCE  and 
ACBn  DEC  ;  then  the  homologous  sides, 
or  the  sides  adjacent  to  the  equal  angles, 
will  be  proportional,  so  that  we  shall 
have  BC  ;  CE  :  :  AB  :  CD  :  :  AC  : 
DE. 

Place  the  homologous  sides  BC,  CE  in  the  same  straight 
line  ;  and  produce  the  sides  BA,  ED,  till  they  meet  in  F. 

Since  BCE  is  a  straight  line,  and  the  angle  BCA  is  equal  to 
CED,  it  follows  that  AC  is  parallel  to  DE  (Book  I.  Prop.  XIX. 
Cor.  2.).  In  like  manner,  since  the  angle  ABC  is  equal  to 
DCE,  the  line  AB  is  parallel  to  DC.  Hence  the  figure  ACDF 
is  a  parallelogram. 

In  the  triangle  BFE,  the  line  AC  is  parallel  to  the  base  FE ; 
hence  we  have  BC  :  CE  :  :  BA  :  AF  (Prop.  XV.);  or  put- 
ting CD  in  the  place  of  its  equal  AF, 

BC  :  CE  :  :  BA  :  CD. 

In  the  same  triangle  BEF,  CD  is  parallel  to  BF  which  may 
be  considered  as  the  base ;  and  we  have  the  proportion 
BC  :  CE  : :  FD  :  DE ;  or  putting  AC  in  the  place  of  its  equal  FD, 

BC  :  CE  :  :  AC  :  DE. 

And  finally,  since  both  these  proportions  contain  the  same 
ratio  BC  :  CE,  we  have 

AC  :  DE  :  :  BA  :  CD. 

Thus  the  equiangular  triangles  BAC,  CED,  have  their  ho- 
mologous sides  proportional.  But  two  figures  are  similar  when 
they  have  their  angles  equal,  each  to  each,  and  their  homolo- 
gous sides  proportional  (Def.  1.) ;  consequently  the  equiangu- 
lar triangles  BAC,  CED,  are  two  similar  figures. 

Cor,  For  the  similarity  of  two  triangles,  it  is  enough  that 
they  have  two  angles  equal,  each  to  each ;  since  then,  the 
third  will  also  be  equal  in  both,  and  the  two  triangles  will  be 
equiangular. 


BOOK  IV.  m 

SchuUum.  Observe,  that  in  similar  triangles,  the  homolo- 
gous sides  arc  opposite  to  the  equal  angles  ;  thus  the  angle  ACB 
being  equal  to  DEC,  the  side  AB  is  homologous  to  DC  ;  in  like 
manner,  AC  and  DE  are  homologous,  because  opposite  to  the 
equal  angles  ABC,  DCE.  When  the  homologous  sides  are  de- 
termined, it  is  easy'to  form  the  proportions : 

AB  :  DC  :  :  AC  :  DE  :  :  BC  :  CE. 


PROPOSITION  XIX.    THEOREM. 

Two  triangles^  which  have  their  homologous  sides  proportional^ 
are  equiangular  and  similar. 

In  the  two  triangles  BAC,  DEF, 
suppose  we  have  BC  :  EF  :  :  AB 
.  DE  :  :  AC  :  DF ;  then  wilj  the 
triangles  ABC,  DEF  have  their  an- 
gles equal,  namely,  A=D,  B=E, 
C=F.  ^ 

At  the  point  E,  make  the  angle 
FEG  =  B,  and  at  F,  the  angle  EFG  =  C ;  the  third  G  will  be 
equal  to  the  third  A,  and  the  two  triangles  ABC,  EFG  will  be 
equiangular  (Book  I.  Prop.  XXV.  Cor.  2.).  Therefore,  by  the 
last  theorem,  we  shall  have  BC  :  EF  :  :  AB  :  EG ;  but,  by 
hypothesis,  we  have  BC  :  EF  :  :  AB  :  DE;  hence  EG =DE. 
By  the  same  theorem,  we  shall  a!so  have  BC  ;  EF  :  :  AC  : 
FG ;  and  by  hypothesis,  we  have  BC  :  EF  :  :  AC  :  DF ; 
hence  FG=:DF.  Hence  the  triangles  EGF,  DEF,  having  their 
three  sides  equal,  each  to  each,  are  themselves  equal  (Book  I. 
Prop.  X.).  But  by  construction,  the  triangles  EGF  and 
ABC  are  equiangular ;  hence  DEF  and  ABC  are  also  equian- 
gular and  similar. 

Scholium  1.  By  the  last  two  propositions,  it  appears  that  in 
triangles,  equahty  among  the  angles  is  a  consequence  of  pro- 
portionality among  the  sides,  and  conversely  ;  so  that  either  of 
those  conditions  sufficiendy  determines  the  similarity  of  two 
triangles.  The  case  is  different  with  regard  to  figures  of 
more  than  three  sides  :  even  in  quadrilaterals,  the  proportion 
between  the  sides  may  be  altered  without  altering  the  angles, 
or  the  angles  may  be  altered  without  altering  the  proportion 
between  the  sides  ;  and  thus  proportionality  among  the  sides 
cannot  be  a  consequence  of  equality  among  the  angles  of  two 
quadrilaterals,  or  vice  versa.     It  is  evident,  for  example,  that 

H 


86  GEOMETRY. 

by  drawing  EF  parallel  to  BC,  the  angles  of 
the  quadrilateral  AEFD,  are  made  equal  to 
those  of  ABCD,  though  the  proportion  be- 
tween the  sides  is  different ;  and,  in  like  man- 
ner, without  changing  the  four  sides  AB,  BC, 
CD,  AD,  we  can  make  the  point  B  approach 
D  or  recede  from  it,  which  will  change  the 
angles. 

Scholium  2.  The  two  preceding  propositions,  which  in  strict- 
ness form  but  one,  together  with  that  relating  to  the  square  of 
the  hypothenuse,  are  the  most  important  and  fertile  in  results 
of  any  in  geometry  :  they  are  almost  sufficient  of  themselves 
for  every  application  to  subsequent  reasoning,  and  for  solving 
every  problem.  The  reason  is,  that  all  figures  may  be  divided 
into  triangles,  and  any  triangle  into  two  right  angled  triangles. 
Thus  the  general  properties  of  triangles  include,  by  implica- 
tion, those  of  all  figures. 


PROPOSITION  XX.    THEOREM.  ; 

Two  triangles,  which  have  an  angle  of  the  one  equal  to  an  angle 
of  the  other,  and  the  sides  containing  those  angles  proportional, 
are  similar. 

In  the  two  triangles  ABC,  DEF,  let 
the  angles  A  and  D  be  equal  ;  then,  if 
AB  :  DE  :  :  AC  :  DF,  the  two  trian- 
gles will  be  similar. 

Take  AG-.DE,  and  draw  GH  paral- 
lel to  BC.  The  angle  AGH  will  be  equal 
to  the  angle  ABC  (Book  I.  Prop.  XX. 
Cor  3.) ;  and  the  triangles  AGH,  ABC,  will  be  equiangular :  \ 
hence  we  shall  have  AB  :  AG  :  :  AC  :  AH.     But  by  hypo- 
thesis, we  have  AB  :  DE  :  :  AC  :  DF ;  and  by  construction, 
AG=DE :  hence  AH=DF.     The  two  triangles  AGH,  DEF, ' 
have  an  equal  angle  included  between  equal  sides ;  therefore  ^ 
they  are  equal  :  but  the  triangle  AGH  is  similar  to  ABC :  there* 

fore  DEF  is  also  similar  to  ABC. 

J 

I 


BOOK  IV 


#7 


PROPOSITION  XXI.    THEOREM. 


Two  triangles,  which  have  their  homologous  sides  parallel,  or 
perpendicular  to  each  other,  are  similar. 


Let  BAG,  EDF,  be  two  triangles. 

First.  If  the  side  AB  is  parallel  to  DE,  and 
BC  to  EF,  the  angle  ABC  will  be  equal  to 
DEF  (Book  I.  Prop.  XXIV.) ;  if  AC  is  parallel 
to  DF,  the  angle  ACB  will  be  equal  to  DFE, 
and  also  BAC  to  EDF ;  hence  the  triangles 
ABC,  DEF,  are  equiangular;  consequently 
they  are  similar  (Prop.  XVIII.). 


Secondly.  If  the  side  ©E  is  perpen- 
dicular to  AB,  and  the  side  DF  to  AC, 
the  two  angles  I  and  H  of  the  quadri- 
lateral AIDH  will  be  right  angles  ;  and 
since  all  the  four  angles  are  together 
equal  to  four  right  angles  (Book  I.  Prop. 
XXVI.  Cor.  l.),the  remaining  two  I  AH, 
IDH,  will  be  together  equal  to  two  right  ^ 
angles.  But  the  two  angles  EDF,  IDH,  are  also  equal  to  two 
right  angles :  hence  the  angle  EDF  is  equal  to  lAH  or  BAC. 
In  like  manner,  if  the  third  side  EF  is  perpendicular  to  the  third 
side  BC,  it  may  be  shown  that  the  angle  DFE  is  equal  to  C,  and 
DEF  to  B  :  hence  the  triangles  ABC,  DEF,  which  have  the 
sides  of  the  one  perpendicular  to  the  corresponding  sides  of  the 
other,  are  equiangular  and  similar. 

Scholium.  In  the  case  of  the  sides  being  parallel,  the  homolo- 
gous sides  are  the  parallel  ones  :  in  the  case  of  their  being  per- 
pendicular, the  homologous  sides  are  the  perpendicular  ones. 
Thus  in  the  latter  case  DE  is  homologous  with  AB,  DF  with 
AC,  and  EF  with  BC. 

The  case  of  the  per[A3ndicular  sides  might  present  a  rela- 
tive position  of  the  two  triangles  different  from  that  exhibited 
in  the  diagram.  But  we  might  always  conceive  a  triangle 
DEF  to  be  constructed  within  the  triangle  ABC,  and  such  that 
its  sides  should  be  parallel  to  those  of  the  triangle  compared 
with  ABC  ;  and  then  the  demonstration  given  in  the  text  would 
apply. 


68 


GEOMETRY. 


PROPOSITION  XXII.    THEOREM. 

In  any  triangle,  if  a  line  he  drawn  parallel  to  the  base,  then,  all 
lines  drawn  from  the  vertex  will  divide  the  base  and  the  pai^- 
allel  into  proportional  parts. 

Let  DE  be  parallel  to  the  base  BC,  and 
the  other  lines  drawn  as  in  the  figure  ; 
then  will 
DI  :  BF  :  :  IK  :  FG  :  :  KL  :  GH. 

For,  since  \)\  is  parallel  to  BF,  the 
triangles  ADI  and  ABF  are  equiangu- 
lar ;  and  we  have  DI  :  BF  :  :  AI  : 
AF ;  and  since  IK  is  parallel  to  FG, 
we  have  in  like  manner  AI  :  AF  :  : 
IK  :  FG ;  hence,  the  ratio  AI  :  AF  being  common,  we  shall 
have  DI  :  BF  :  :  IK  :  FG.  In  the  same  manner  we  shall 
find  IK  :  FG  :  :  KL  ;  GH  ;  and  so  with  the  other  segments ; 
hence  the  line  DE  is  divided  at  the  points  I,  K,  L,  in  the  same 
proportion,  as  the  base  BC,  at  the  points  F,  G,  H. 

Cor.  Therefore  if  BC  were  divided  into  equal  parts  at  the 
points  F,  G,  H,  the  parallel  DE  would  also  be  divided  into  equal 
parts  at  the  points  1,  K,  L. 


PROPOSITION  XXIII.    THEOREM. 


If  from  the  right  angle  of  a  right  angled  triangle,  a  peiyendicu- 
lar  he  let  fall  on  the  hypothenuse  ;  then, 

\st.  The<  two  partial  triangles  thus  formed,  will  he  similar  to  each 
other,  and  to  the  whole  triangle. 

2d.  Either  side  including  the  right  angle  will  he  a  mean  propor- 
tional hetiveen  the  hypothenuse  and  the  adjacent  segment. 

*Sd.  The  perpendicular  will  be  a  mean  proportional  between  the 
two  segments  of  the  hypothenuse. 

Let  BAC  be  a  right  angled  triangle,  and  AD  perpendicular 
to  the  hypothenuse  BC. 

First.  The  triangles  BAD  and  BAC 
have  the  common  angle  B.  the  right 
angle  BDA=BAC,  and  therefore  the 
third  angle  BAD  of  the  one,  equal  to 
the  third  angle  C,  of  the  other  (Book 
I.  Prop.  XXV.  Cor  2.) :  hence  those 
two  triangles  are   equiangular    and 


BOOK  IV.  89 

similar.  In  the  same  manner  it  may  be  shown  that  the  trian- 
gles DAC  and  BAG  are  similar ;  hence  all  the  triangles  are 
equiangular  and  similar. 

Secondly.  The  triangles  BAD,  BAC,  being  similar,  their 
homologous  sides  are  proportional.  But  BD  in  the  small  tri- 
angle, and  BA  in  the  large  one,  are  homologous  sides,  because 
they  lie  opposite  the  equal  angles  BAD,  BCA ;  the  hypothe- 
nuse  BA  of  the  small  triangle  is  homologous  with  the  hypo- 
thenuse  BC  of  the  large  triangle  :  hence  the  proportion  BD  : 
BA  :  :  BA  :  BC.  By  the  same  reasoning,  we  should  find 
DC  :  AC  :  :  AC  :  BC  ;  hence,  each  of  the  sides  AB,  AC,  is 
a  mean  proportional  between  the  hypothenuse  and  the  segment 
adjacent  to  that  side. 

Thirdly.  Since  the  triangles  ABD,  ADC,  are  similar,  by 
comparing  their  homologous  sides,  we  have  BD  :  AD  ;  :  AD 
:  DC  ;  hence,  the  perpendicular  AD  is  a  mean  proportional 
between  the  segments  BD,  DC,  of  the  hypothenuse. 

Scholium.  Since  BD  :  AB  :  :  AB  :  BC,  the  product  of  the 
extremes  will  be  equal  to  that  of  the  means,  or  AB^=:BD.BC. 
For  the  same  reason  we  have  AC^— DC.BC  ;  therefore  AB^+ 
AC2=BD.BC  +  DC.BC=  (BD  +  DC).BC=BC.BC=BC2;  or 
the  square  described  on  the  hypothenuse  BC  is  equivalent  to 
the  squares  described  on  the  two  sides  AB,  AC.  Thus  we  again 
arrive  at  the  property  of  the  square  of  the  hypothenuse,  by  a 
path  very  diflferent  from  that  which  formerly  conducted  us  to 
it :  and  thus  it  appears  that,  strictly  speaking,  the  property  of 
the  square  of  the  hypothenuse,  is  a  consequence  of  the  more 
general  property,  that  the  sides  of  equiangular  triangles  are 
proportional.  Thus  the  fundamental  propositions  of  geometry 
are  reduced,  as  it  were,  to  this  single  one,  that  equiangular  tri- 
angles have  their  homologous  sides  proportional. 

It  happens  frequently,  as  in  this  instance,  that  by  deducing 
consequences  from  one  or  more  propositions,  we  are  led  back 
to  some  proposition  already  proved.  In  fact,  the  chief  charac- 
teristic of  geometrical  theorems,  and  one  indubitable  proof  of 
their  certainty  is,  that,  however  we  combine  them  together, 
provided  only  our  reasoning  be  correct,  the  results  we  obtain 
are  always  perfectly  accurate.  The  case  would  be  different, 
if  any  proposition  were  false  or  only  approximately  true  :  it 
would  frequently  happen  that  on  combining  the  propositions 
together,  the  error  would  increase  and  become  perceptible. 
Examples  of  this  are  to  be  seen  in  all  the  demonstrations,  in 
which  the  reductio  ad  absurdum  is  employed.  In  such  demon- 
strations, where  the  object  is  to  show  that  two  quantities  are 
equal,  we  proceed  by  showing  that  if  there  existed  the  smallest 

H* 


90 


GEOMETRY. 


inequality  between  the  quantities,  a  train  of  accurate  reason- 
ing would  lead  us  to  a  manifest  and  palpable  absurdity;  from 
which  we  are  forced  to  conclude  that  the  two  quantities  are 
equal. 

Cor.  If  from  a  point  A,  in  the  circumference 
of  a  circle,  two  chords  AB,  AC,  be  drawn  to 
the  extremities  of  a  diameter  BC,  the  triangle 
BAG  will  be  right  angled  at  A  (Book  III.  Prop.  ^  J^  C 

XVIII.  Gor.  2.) ;  hence,  first,  the  perpendicular  AD  is  a  mean 
proportional  between  the  two  segments  BD,  DG,  of  the  diameter^ 
or  what  is  the  same,  AD^=BD.DG. 

Ilence  also,  in  the  second  place,  the  chord  AB  is  a  mean  pro- 
portional between  the  diameter  BC  and  the  adjacent  segment  BD, 
or,  what  is  the  same,  AB^nBD.BC.  In  like  manner,  we  have 
AC^=GD.BG  ;  hence  AB^  :  AC^  :  :  BD  :  DC  :  and  com- 
paring AB'^  and  AC^,  to  BC^,  we  have  AB^  :  BC^ :  :  BD  : 
BC,  and  AC^  :  BC^  :  :  DC  :  BC.  Those  proportions  between 
the  squares  of  the  sides  compared  with  each  other,  or  with  the 
square  of  the  hypothenuse,  have  already  been  given  in  the  third 
and  fourth  corollaries  of  Prop.  XI. 


PROPOSITION  XXIV.    THEOREM. 


Two  triangles  having  an  angle  in  each  equals  are  to  each  other 
as  the  rectangles  of  the  sides  which  contain  the  equal  angles. 

In  the  two  triangles  ABC,  ADE,  let  the  angle  A  be  equal  to 
the  angle  A ;  then  will  the  triangle 

ABC  :  ADE  :  :  AB.AC  :  AD.AE. 

Draw  BE.  The  triangles 
ABE,  ADE,  having  the  com- 
mon vertex  E,  have  the  same 
altitude,  and  consequently  are 
to  each  other  as  their  bases 
(Prop.  VI.  Cor.) :  that  is, 

ABE  :  ADE  :  :  AB  :  AD. 


In  like  manner, 

ABC 


ABE 


AC  :  AR 


Multiply  together  the  corresponding  terms  of  these  proportions, 
omitting  the  common  term  ABE ;  we  have 

ABC  :  ADE  :  AB.AC  :  AD.AE. 


BOOK  IV.  01 

Cor.  Jlcnce  the  two  triangles  would  be  equivalent,  if  the 
rectangle  AB.AC  were  equal  to  the  rectangle  AD.AE,  or  if 
we  had  AB  :  AD  :  :  AE  :  AC  i  which  would  happen  if  DC 
were  paraliv^.l  to  BE. 


PROPOSITION  XXV.    THEOREM. 

Two  similar  triangles  are  to  each  other  as  the  squares  described 
on  their  homologous  sides. 

Let  ABC,  DEF,  be  two  similar  trian- 
gles, having  the  angle  A  equal  to  D,  and 
the  angle  B=E. 

Then,  first,  by  reason  of  the  equal  an- 
gles A  and  D,  according  to  the  last  pro- 
position, we  shall  have 

ABC  :  DEF  :  :  AB.AC  :  DE.DF. 
Also,  because  the  triangles  are  similar, 

AB  :  DE  :  :  AC  :  DF, 
And  multiplying  the  terms  of  this  proportion  by  the  corres- 
ponding terms  of  the* identical  proportion, 

AC  :  DF  :  :  AC  :  DF, 
there  v/ill  result  • 

AB.AC  :  DE.DF  :  :  AC^  :  DP. 

Consequently, 

ABC  :  DEF  :  :  AC^  :  DF^. 

Therefore,  two  similar  triangles  ABC,  DEF,  are  to  each 
other  as  the  squares  described  on  their  homologous  sides  AC, 
DF,  or  as  the  squares  of  any  other  two  homologous  sides. 


PROPOSITION  XXVI.    THEOREM. 

7\vo  similar  polygons  are  composed  of  the  same  number  of  tri- 
angles ^  similar  each  to  each,  and  similarly  situated. 


92  GEOMETRY. 

Let  ABCDE,  FGHIK,  be  two  similar  polygons. 

From  any  angle  A,  in  c 

the  polygon  ABCDE, 
draw  diagonals  AC,  AD 
to  the  other  angles.  From 
the  homologous  angle  F, 
in  the  other  polygon 
FGHIK,  draw  diagonals 
FH,  FI  to  the  other  an- 
gles. 

These  polygons  being  similar,  the  angles  ABC,  FGH,  which 
are  homologous,  must  be  equal,  and  the  sides  AB,  BC,  must 
also  be  proportional  to  FG,  GH,  that  is,  AB  :  FG  :  :  BC  : 
GH  (Def.  1.).  Wherefore  the  triangles  ABC,  FGH,  have  each 
an  equal  angle,  contained  between  proportional  sides  ;  hence 
they  are  similar  (Prop.  XX.) ;  therefore  the  angle  BCA  is  equal 
to  GHF.  Take  away  these  equal  angles  from  the  equal  angles 
BCD,  GHI,  and  there  remains  ACD==FHI.  But  since  the 
triangles  ABC,  FGH,  are  similar,  we  have  AC  :  FH  :  :  BC  : 
GH ;  and,  since  the  polygons  are  similar,  BC  :  GH  :  :  CD  : 
HI ;  hence  AC  :  FH  :  :  CD  :  HI.  But  the  angle  ACD,  we 
already  know,  is  equal  to  FHI ;  hence  the  triangles  ACD,  FHI, 
have  an  equal  angle  in  each,  included  between  proportional 
sides,  and  are  consequently  similar  (Prop.  XX.).  In  the  same 
manner  it  might  be  shown  that  all  the  remaining  triangles  are 
similar,  whatever  be  the  number  of  sides  in  the  polygons  pro- 
posed :  therefore  two  similar  polygons  are  composed  of  the 
same  number  of  triangles,  similar,  and  similarly  situated. 

Scholium,  The  converse  of  the  proposition  is  equally  true  : 
If  two  polygons  are  composed  cf  the  same  number  of  triangles 
similar  and  similarly  situated,  those  two  polygons  will  be  similar. 

For,  the  similarity  of  the  respective  triangles  will  give  the 
angles,  ABC=FGH, BCA^GHF,  ACD=FHI :  hence  BCD= 
GHI,  likewise  CDE=HIK,  &c.  Moreover  we  shall  have 
AB  :  FG  :  :  BC  :  GH  :  :  AC  :  FH  :  :  CD  :  HI,&c.;  hence 
the  two  polygons  have  their  angles  equal  and  their  sides  pro- 
portional ;  consequently  they  are  similar. 


PROPOSITION  XXVII.    THEOREM. 

The  conto^irs  or  perimeters  of  similar  polygons  are  to  each  other 
as  the  homologous  sides :  and  the  areas  are  to  each  other  as 
the  squares  described  on  those  sides. 


BOOK  IV.  03 

First.  Since,  by  the 
nature  of  similar  figures, 
we  have  AB  :  FG  :  : 
BC  :  GH  :  :  CD  ;  HI, 
&c.  we  conclude  from 
this  series  of  equal  ratios 
that  the  sum  of  the  ante- 
eecieuts  AB  +  BC  +  CD, 
&c.,  which  makes  up  the  perimeter  of  the  first  polygon,  is  to 
the  sum  of  the  consequents  FG  +  GH  +  HI,  &c.,  which  makes 
up  the  perimeter  of  the  second  polygon,  as  any  one  antecedent 
is  to  its  consequent ;  and  therefcjre,  as  the  side  AB  is  to  its  cor- 
responding side  FG  (Book  II.  Prop.  X.). 

Secondly.  Since  the  triangles  ABC,  FGH  are  similar,  we 
shall  have  the  triangle  ABC  :  FGH  :  :  AC^  :  FH^  (Prop. 
XXV.) ;  and  in  like  manner,  from  the  similar  triangles  ACD, 
FHI,  we  shall  have  ACD  :  FHI  :  :  AC^  :  FH^;  therefore,  by 
reason  of  the  common  ratio,  AC^  :  FH^,  we  have 

ABC  :  FGH  :  :  ACD  :  FHI. 
By  the  same  mode  of  reasoning,  we  should  find 

ACD  :  FHI  :  :  ADE  :  FlK; 
and  so  on,  if  there  were  more  triangles.  And  from  this  series 
of  equal  ratios,  we  conclude  that  the  sum  of  the  antecedents 
ABC  +  ACD  +  ADE,  or  the  polygon  ABCDE,  is  to  the  sum  cf 
the  consequents  FGH  +  FHI  +  FIK,  or  to  the  polygon  FGHIK, 
as  one  antecedent  ABC,  is  to  its  consequent  FGH,  or  as  AB^ 
is  to  FG-  (Prop.  XXV.)  ;  hence  the  areas  of  similar  poly- 
gons are  to  each  other  as  the  squares  described  on  the  homolo- 
gous sides. 

Cor.  If  three  similar  figures  were  constructed,  on  the 
three  sides  of  a  right  angled  triangle,  the  figure  on  the  hypo- 
thenuse  would  be  equivalent  to  the  sum  of  the  other  two  :  for 
the  three  figures  are  proportional  to  the  squares  of  their 
homologous  sides ;  but  the  square  of  the  hypothenuse  is 
equivalent  to  the  sum  of  the  squares  of  the  two  other  sides ; 
hence,  &c. 


PROPOSITION  XXVIII.    THEOREM. 

The  segments  of  two  chords,  which  intersect  each  other  vn  a  circle, 
are  reciprocally  proportional. 


u 


GEOMETRY. 


Let  the  chords  AB  and  CD  intersect  at  O :  then  will 
AO  :  DO  :  :  OC  :  OB. 

Draw  AC  and  BD.  In  the  triangles  ACO, 
BOD,  the  angles  at  O  are  equal,  being  verti- 
cal ;  the  angle  A  is  equal  to  the  angle  D,  be- 
cause both  are  inscribed  in  the  same  segment 
(Book  III.  Prop.  XVIII.  Cor.  1.) ;  for  the  same 
reason  the  angle  C— B ;  the  triangles  are  there- 
fore similar,  and  the  homologous  sides  give  the  proportion 
AO  :  DO  :  :  CO  :  OB. 

Cor.  Therefore  AO.OB=DO.CO:  hence  the  rectangle 
under  the  two  segments  of  the  one  chord  is  equal  to  the  rect- 
angle under  the  two  segments  of  the  other. 


PROPOSITION  XXIX.    THEOREM. 

If  from  the  same  point  without  a  circle,  two  secants  he  drawn 
terminating  in  the  concave  arc,  the  whole  secants  will  he  recip' 

srannJIg/  2^r'e?po<rii^9^ty1  i-»  i?ioit'  eJCtcmul  segments. 


Let  the  secants  OB,  OC,  be  drawn  from  the  point  O 
then  will 

OB  :  OC  :  :  OD  :  OA. 
For,  drawing  AC,  BD,  the  triangles  OAC, 
OBD  have  the  angle  O  common  ;  likewise  the 
angle  B=C  (Book  IIL  Prop.  XYIII.Cor.  1.); 
these  triangles  are  therefore  similar ;  and  their 
homologous  sides  give  the  proportion, 
OB  :  OC  :  :  OD  :  OA. 

Cor.     Hence  the  rectangle  OA.OB  is  equal 
to  the  rectangle  OC.OD.  B 

Scholium,  This  proposition,  it  may  be  observed,  bears  a 
great  analogy  to  the  preceding,  and  differs  from  it  only  as  the 
two  chords  AB,  CD,  instead  of  intersecting  each  other  within, 
cut  each  other  without  the  circle.  The  following  proposition 
may  also  be  regarded  as  a  particular  case  of  the  proposition 
just  demonstrated. 


BOOK  IV. 


95 


PROPOSITION   XXX.    THEOREM. 


If  from  the  same  point  without  a  circle ,  a  tangent  and  a  secant 
be  drawn,  the  tangent  will  he  a  mean  proportional  between  the 
secant  and  its  external  segment. 

From  the  point  O,  let  the  tangent  OA,  and  the  secant  OC  be 
be  drawn  ;  then  will 

OC  :  OA  :  :  OA  :  OD,  or  OA^^OC.OD. 
For,  drawing  AD  and  AC,  the  triangles  O 
OAD,  OAC,  have  the  angle  O  common;  also 
the  angle  OAD,  formed  by  a  tangent  and  a 
chord,  has  for  its  measure  half  of  the  arc  AD 
(Book  III.  Prop.  XXI.) ;  and  the  angle  C  has 
the  same  measure :  hence  the  angle  OAD  = 
C ;  therefore  the  two  triangles  are  similar, 
and  we  have  the  proportion  OC  :  OA  :  : 
AO  :  OD,  which  gives  OA'-=OC.OD. 

X 


PROPOSITION  XXXI.    THEOREM. 

If  either  angle  of  a  triangle  he  bisected  by  a  line  terminating  in 
the  opposite  side,  the  rectangle  of  the  sides  including  the  bi- 
sected angle,  is  equivalent  to  the  square  of  the  bisecting  line 
together  with  the  rectangle  contained  by  the  segments  of  the 
third  side. 


In  the  triangle  BAC,  let  AD  bisect  the  angle  A ;  then  will 
AB.AC=:ADHBD.DC. 

Describe  a  circle  through  the  three  points 
A,  B,  C  ;  produce  AD  till  it  meets  the  cir- 
cumference, and  draw  CE. 

The  triangle  BAD  is  similar  to  the  trian- 
gle EAC ;  for,  by  hypothesis,  the  angle 
BAD  =  EAC;  also  the  angle  iB=E,' since 
they  are  both  measured  by  half  of  the  arc 
AC  ;  hence  these  triangles  are  similar,  and 
the  homologous  sides  give  the  proportion  BA  :  AE  :  :  AD  : 
AC  ;  hence  BA.AC=AE.AD  ;  but  AE=AD  +  DE,  and  multi- 
plying each  of  these  equals  by  AD,  we  have  AE.AD=:AD'H 
Ab.DE;  now  AD.DE=BD.DC  (Prop.  XXVIII.)  ;  hence, 
finallv, 

BA.AC==AD2+BD.DC. 


m 


OG 


GEOMETIIY. 


PROPOSITION  XXXII.    THEOREM. 

In  every  triangle,  the  rectangle  contained  hij  two  sides  is  equiva- 
lent to  the  rectangle  contained  by  the  diameter  of  the  circum- 
scribed circle^  and  the  perpendicular  let  fall  upon  the  third 
side. 


In  the  triangle  ABC,  let  AD  be  drawn  perpendicular  to  BC  ; 
and  let  EC  be  the  diameter  of  the  circumscribed  circle  :  then 
will 

AB.AC=AD.CE. 

For,  drawing  AE,  the  triangles  ABD, 
AEC,  are  right  angled,  tiie  one  at  D,  the 
other  at  A:  also  the  angle  B  — E ;  these  tri- 
angles are  therefore  similar,  and  they  give 
the  proportion  AB  :  CE  :  :  AD  :  AC ;  and 
hence  AB.ACrr:CE.AD. 

Cor.  If  these  equal  quantities  be  multiplied  by  the  same 
quantity  BC,  there  will  result  AB.AC.BC  =  CE.AD.BC  ;  now 

AD.BC  is  double  of  the  area  of  the  triangle  (Prop.  VI.) ;  there- 
fore the  product  of  three  sides  of  a  triangle  is  equal  to  its  area 
multiplied  by  twice  the  diameter  of  the  ci?xumscribed  circle. 

The  product  of  three  lines  is  sometimes  called  a  solid,  for  a 
reason  that  shall  be  seen  afterwards.  Its  value  is  easily  con- 
ceived, by  imagining  that  the  lines  are  reduced  into  numbers, 
and  multiplying  these  numbers  together. 

Scholium.  It  may  also  be  demonstrated,  that  the  area  of 
a  triangle  is  equal  to  its  perimeter  multiplied  by  half  the  radius 
of  the  inscribed  circle. 

For,  the  triangles  AOB,  BOC, 
AOC,  which  have  a  common 
vertex  at  O,  have  for  their  com- 
mon altitude  the  radius  of  the 
inscribed  circle  ;  hence  the  sum 
of  these  triangles  will  be  equal 
to  the  sum  of  the  bases  AB,  BC, 
AC,  multiplied  by  half  the  radius 

OD ;  hence  the*^  area  of  the  triangle  ABC  is  equal  to  the 
perimeter  multiplied  by  half  the  radius  of  the  inscribed  circle. 


BOOK  IV. 


97 


PROPOSITION  XXXIII.    THEOREM. 


In  every  quadrilateral  inscribed  in  a  circle,  the  rectangle  of  the 
two  diagonals  is  equivalent  to  the  sum  of  the  rectangles  of  the 
opposite  sides. 

In  the  quadrilateral  ABCD,  we  shall  have 
AC.BD=AB.CD+AD.BC. 

Take  the  arc  CO = AD,  and  draw  BO 
meeting  the  diagonal  AC  in  I. 

The  angle  ABD=CBI,  since  the  one 
has  for  its  measure  half  of  the  arc  AD, 
and  the  other,  half  of  CO,  equal  to  AD  ; 
the  angle  ADB=BCI,  because  they  are 
both  inscribed  in  the  same  segment 
AOB  ;  hence  the  triangle  ABD  is  similar 
to  the  triangle  IBC,  and  we  have  the 
proportion  AD  ;  CI  :  :  BD  :  BC ;  hence  AD.BC=:CI.BD. 
Again,  the  triangle  ABl  is  similar  to  the  triangle  BDC  ;  for  the 
arc  AD  being  equal  to  CO,  if  OD  be  added  to  each  of  them, 
we  shall  have  the  arc  AO=r:DC  ;  hence  the  angle  ABI  is  equal 
to  DBC  ;  also  the  angle  BAI  to  BDC,  because  they  are  in- 
scribed in  the  same  segment ;  hence  the  triangles  ABI,  DBC, 
are  similar,  and  the  homologous  sides  give  the  proportion  AB  : 
BD  :  :  AI  :  CD ;  hence  AB.CD=AI.BD. 

Adding  the  two  results  obtained,  and  observing  that 

AI.BD  +  CI.BD = (AI  +  CI).BD=:AC.BD, 
we  shall  have 

AD.BC+AB.CD=AC.BD. 


OS 


GEOMETRY. 


^t 


PROBLEMS  RELATING  TO  THE  FOURTH  BOOK. 


PROBLEM  I. 

To  divide  a  given  straight  line  into  any  number  of  equal  parts, 
or  into  parts  proportional  to  given  lines. 

First.  Let  it  be  proposed  to  divide  the  line 
AB  into  five  equal  parts.  Through  the  ex- 
tremity A,  draw^  the  indefinite  straight  line 
AG ;  and  taking  AC  of  any  magnitude,  apply 
it  five  times  upon  AG  ;  join  the  last  point 
of  division  G,  and  the  extremity  B,  by  the 
straight  line  GB  ;  then  draw  CI  parallel  to 
GB :  AI  will  be  the  fifth  part  of  the  line 
AB  ;  and  thus,  by  applying  AI  five  times 
upon  AB,  the  line  AB  will  be  divided  into 
five  equal  parts. 

For,  since  CI  is  parallel  to  GB,  the  sides  AG,  AB,  are  cut 
proportionally  in  C  and  I  (Prop.  XV.).  But  AC  is  the  fifth 
part  of  AG,  hence  AI  is  the  fifth  part  of  AB,  ^ 

Secondly.  Let  it  be  pro- 
posed to  divide  the  line  AB 
mto  parts  proportional  to 
the  given  lines  P,  Q,  R. 
Through  A,  draw  the  indefi- 
nite hne  AG  ;  make  AC  = 
P,  CD=Q,  DE:=R;  join 
the  extremities  E  and  B  ; 
and   through    the  points  C, 

D,  draw  CI,  DF,  parallel  to  EB ;  the  line  AB  will  be  divided 
into    parts  AI,  IF,  FB,  proportional    to    the  given  lines  P, 

Q,  R. 

For,  by  reason  of  the  paraLcls  CI,  DF,  EB,  the  parts  AI, 
IF,  FB,  are  proportional  to  the  parts  AC,  CD,  DE ;  and  by 
construction,  these  are  equal  to  the  given  lines  P,  Q,  R. 


BOOK  IV. 


PROBLEM  II. 

To  find  a  fourth  proportional  to  three  given  lineSy  A,  B,  C. 

Draw  the  two  indefi- 
nite lines  DE,  DF,  form- 
ing any  angle  with  each 
other.  Upon  DE  take 
DA=A,  and  DB=B ; 
upon  DF  take  DC=C  ; 
draw  AC  ;  and  through 
the  point  B,  draw  BX 
parallel  to  AC  ;  DX  will  be  the  fourth  proportional  required  ; 
for,  since  BX  is  parallel  to  AC,  we  have  the  proportion 
DA  :  DB  :  :  DC  :  DX ;  now  the  first  three  terms  of  this  pro- 
portion are  equal  to  the  three  given  lines  :  consequently  DX  is 
the  fourth  proportional  required. 

Cor.  A  third  proportional  to  two  given  lines  A,  B,  may  be 
found  in  the  same  manner,  for  it  will  be  the  same  as  a  fourth 
proportional  to  the  three  lines  A,  B,  B. 

PROBLEM  in. 
To  find  a  mean  proportional  between  two  given  lines  A  and  B. 

Upon  the  indefinite  line  DF,  take 
DE=A,  and  EF— B  :  upon  the  whole 
line  DF,  as  a  diameter,  describe  the 
semicircle  DGF ;  at  the  point  E, 
erect  upon  tl}e  diameter  the  perpen- 
dicular EG  meeting  the  circumfe- 
rence in  G ;  EG  will  be  the  mean 
proportional  required. 

For,  the  perpendicular  EG,  let  fall  from  a  point  in  the  cir- 
cumference upon  the  diameter,  is  a  mean  proportional  between 
DE  EF,  the  two  segments  of  the  diameter  (Prop.  XXIII. 
Co  I .) ;  and  these  segments  are  equal  to  the  given  lines  A 
and  B. 


Ai 1 


PROBLEM  IV. 


To  divide  a  given  line  into  two  parts,  such  that  the  greater  part 
shall  be  a  mean  proportional  between  the  whole  line  and  the 
other  part. 


100 


GEOMETRY. 


Let  AB  be  the  given  line. 

At  the  extremity  B  of  the  line 
AB,  erect  the  perpendicular  BC 
equal  to  the  half  of  AB  ;  from  the 
point  C,  as  a  centre,  with  the  ra- 
dius CB,  describe  a  semicircle  ; 
draw  AC  cutting  the  circumfe- 
rence in  D  ;  and  take  AF=AD  :  A  :f 
the  line  AB  will  be  divided  at  the  point  F  in  the  manner  re- 
quired ;  that  is,  we  shall  have  AB  :  AF  :  :  AF  :  FB. 

For,  AB  being  perpendicular  to  the  radius  at  its  extremity, 
is  a  tangent ;  and  if  AC  be  produced  till  it  again  meets  the 
circumference  in  E,  we  shall  have  AE  :  AB  :  :  AB  :  AD 
(Prop.  XXX.) ;  hence,  by  division,  AE— AB  :  AB  :  :  AB— 
AD  :  AD.  But  since  the  radius  is  the  half  of  AB,  the  diame- 
ter DE  is  equal  to  AB,  and  consequently  AE-t-AB= AD=AF  ; 
also,  because  AF=AD,  we  have  AB — ^AD=r:FB ;  hence 
AF  :  AB  :  :  FB  :  AD  or  AF ;  whence,  by  exchanging  the 
extremes  for  the  means,  AB  :  AF  :  :  AF  :  FB. 


Scholium,  This  sort  of  division  of  the  line  AB  is  called  di 
vision  in  extreme  and  mean  ratio  :  the  use  of  it  will  be  per- 
ceived in  a  future  part  of  the  work.  It  may  further  be 
observed,  that  the  secant  AE  is  divided  in  extreme  and  mean 
ratio  at  the  point  D  ;  for,  since  AB=DE,  we  have  AE  :  DE 
:  :  DE  :  AD. 


PROBLEM  V. 

Through  a  given  point,  in  a  given  angle,  to  draw  a  line  so  that 
the  segments  comprehended  between  the  point  and  the  two  sides 
of  the  angle,  shall  be  equal. 

Let  BCD  be  the  given  angle,  and  A  the  given  point. 

Through  the  point  A,  draw  AE  paral- 
lel to  CD,  make  BE=CE,  and  through 
the  points  B  and  A  draw  BAD ;  this  will 
be  the  line  required. 

For,  AE  being  paralV!  +0  CD,  we  have 
BE  :  EC  :  :  BA  :  AD ,  W*  B^.^TC  ; 
therefore  BA=AD. 


BOOK  IV. 

PROBLEM  VI. 


101 


To  describe  a  square  that  shall  he  equivalent  to  a  given  paralleh' 
granif  or  to  a  given  triangle. 


First,  Let  ABCD  be 
the  given  parallelogram, 
AB  its  base,  DE  its  alti- 
tude ;  between  AB  and 
DE  find  a  mean  propor- 
tional XY ;  then  will  the 
square    described     upon 


E 


XY  be  equivalent  to  the  parallelogram  ABCD. 

For,  by  construction,  AB  ;  XY  :  :  XY  :  DE ;  therefore, 
XY2= AB.DE ;  but  x\B.DE  is  the  measure  of  the  parallelogram, 
and  XY^  that  of  the  square  ;  consequently,  they  are  equiva- 
lent. 

Secondly,  Let  ABC  be  the 
given  triangle,  BC  its  base, 
AD  its  altitude  :  find  a  mean 
proportional  between  BC  and 
the  half  of  AD,  and  let  XY  be 
that  mean ;  the  square  de- 
scribed upon  XY  will  be  equi- 
valent to  the  triangle  ABC. 

For,  since  BC  :  XY  :  :  XY  :  lAD,  it  follows  that  XY^^ 
BC.iAD  ;  hence  the  square  described  upon  XY  is  equivalent 
to  the  triangle  ABC. 


PROBLEM  VII. 

Upon  a  given  line,  to  describe  a  rectangle  that  shall  he  equiva- 
lent to  a  given  rectangle. 

Let  AD  be  the  line,  and  ABFC  the  given  rectangle. 

Find  a  fourth  propor 
tional  to  the  three  lines 
AD,  AB,  AC,  and  let  AX 
be  that  fourth  propor- 
tional ;  a  rectangle  con- 
structed with  the  lines 
AD  and  AX  will  be  equi- 
valent to  the  rectangle  ABFC. 

For,  since  AD  :  AB  :  :  AC  :  AX,  it  follows  that  AD. AX = 
AB.AC  ;  hence  the  rectangle  ADEX  is  equivalent  to  the  rect- 
angle ABFC. 


102  GEOMETRY. 

PROBLEM  VIII. 

To  find  two  lines  whose  ratio  shall  he  the  same  as  the  ratio  of 
two  rectangles  contained  by  given  lines. 

Let  A.B,  CD,  be  the  rectangles  contained  by  the  given  Hnes 
A,B,  Candl). 

Find  X,  a  fourth  proportional  to  the  three 
lines  B,  C,  D ;  then  will  the  two  lines  A  and 
X  have  the  same  ratio  to  each  other  as  the 
rectangles  A.B  and  CD. 

For,  since  B  :  C  :  :  D  :  X,  it  follows  that 
CD=B.X;  hence  A.B  :  CD  :  :  A.B  :  B.X 
:  :  A  :  X. 

Cor,  Hence  to  obtain  the  ratio  of  the  squares  described 
upon  the  given  lines  A  and  C,  find  a  third  proportional  X  to 
the  lines  A  and  C,  so  that  A  :  C  :  :  C  :  X;  you  will  then 
have    , 

A.Xrz:C2,  or  A2.X=A.C2 ;  hence 
A2  :  C2  :  :  A  :  X. 

PROBLEM  IX. 

To  find  a  triangle  that  shall  be  equivaleik  to  a  given  polygon. 

Let  ABCDE  be  the  given  polygon. 
Draw  first  the  diagonal  CE  cutting  off 
the  triangle  CDE ;  through  the  point 
D,  draw  DF  parallel  to  CE,  and  ipeet- 
ing  AE  produced  ;  draw  CF:  the  poly- 
gon ABCDE  will  be  equivalent  to  the 
polygon  ABCF,  which  has  one  side 
less  than  the  original  polygon. 

For,  the  triangles  CDE,  CFE,  have  the  base  CE  common, 
they  have  also  the  same  altitude,  since  their  vertices  D  and  F, 
are  situated  in  a  Hne  DF  parallel  to  the  base  :  these  triangles  are 
therefore  equivalent  (Prop.  II.  Cor.  2.).  Add  to  each  of  them 
the  figure  ABCE,  and  there  will  result  the  polygon  ABCDE, 
equivalent  to  the  polygon  ABCF. 

The  angle  B  may  in  like  manner  be  cut  off,  by  substituting 
for  the  triangle  AfiC  the  equivalent  triangle  AGC,  and  thus 
the  pentagon  ABCDE  will  be  changed  into  an  equivalent  tri- 
angle GCF. 

The  same  process  may  be  applied  to  every  other  figure  ; 
for,  by  successively  diminishing  the  number  of  its  sides,  one 
being  retrenched  at  each  step  of  the  process,  the  equivalent 
triangle  will  at  last  be  found. 


BOOK  IV.  108 

Scholium.  We  have  already  seen  that  every  triangle  may 
be  changed  into  an  equivalent  square  (Prob.  VI.) ;  and  thus  a 
square  may  always  be  found  equivalent  to  a  given  rectilineal 
figure,  which  operation  is  called  squaring  the  rectilineal  fiigure, 
or  finding  the  quadrature  of  it. 

The  problem  o{  the  quadrature  of  the  circle^  consists  in  find- 
ing a  square  equivalent  to  a  circle  whose  diameter  is  given. 


PROBLEM  X. 

To  find  the  side  of  a  square  which  shall  he  equivalent  to  the  sum 
or  the  difference  of  two  given  squares. 

Let  A  and  B  be  the  sides  of  the 
given  squares. 

First.  If  it  is  required  to  find  a 
square  equivalent  to  the  sum  of 
these  squares,  draw  the  two  indefi- 
nite lines  ED,  EF,  at  right  angles 
to  each  other;  take  ED = A,  and 
EG=B;  draw  DG:  this  will  be  the  side  of  the  square  re- 
quired, f 

For  the  triangle  DEG  being  right  angled,  the  square  de- 
scribed upon  DG  is  equivalent  to  the  sum  of  the  squares  upon 
ED  and  EG. 

Secondly,  If  it  is  required  to  find  a  square  equivalent  to  the 
difference  of  the  given  squares,  form  in  the  same  manner  the  right 
angle  FEH ;  take  GE  equal  to  the  shorter  of  the  sides  A  and 
B ;  from  the  point  G  as  a  centre,  with  a  radius  GH,  equal  to 
the  other  side,  describe  an  arc  cutting  EH  in  H  :  the  square 
described  upon  EH  will  be  equivalent  to  the  difference  of  the 
squares  described  upon  the  lines  A  and  B. 

For  the  triangle  GEH  is  right  angled,  the  hypothenuse 
GH=:A,  and  the  side  GE=B;  hence  the  square  described 
upon  EH,  is  equivalent  to  the  difference  of  the  squares  A 
and  B. 

Scholium.  A  square  may  thus  be  found,  equivalent  to  the 
sum  of  any  number  of  squares ;  for  a  similar  construction  which 
reduces  two  of  them  to  one,  will  reduce  three  of  them  to  two, 
and  these  two  to  one,  and  so  of  others.  It  would  be  the  same, 
if  any  of  the  squares  were  to  be  subtracted  from  the  sum  of 
the  others. 


104 


GEOMETRY. 


PROBLEM   XI. 


To  find  a  square  which  shall  he  to  a  given  squai  e  as  a  given 
line  to  a  given  line. 

Let  AC  be  the  given 
square,  and  M  and  N  the 
given  lines. 

Upon  the  indefinite 
line  EG,  take  EF=M, 
and  FG=N  ;  upon  EG 
as  a  diameter  describe 
a  semicircle,  and  at  the  point  F  erect  the  perpendicular  FH. 
From  the  point  H,  dravvr  the  chords  HG,  HE,  which  produce 
indefinitely :  upon  the  first,  take  HK  equal  to  the  side  AB  of 
the  given  square,  and  through  the  point  K  draw  KI  parallel  to 
EG ;  HI  will  be  the  side  of  the  square  required. 

For,  by  reason  of  the  parallels  KI,  GE,  we  have  HI  :  HK 
:  :  HE  :  HG;  hence,  HP  :  HK^  :  :  HE^  :  HG^:  but  in  the 
right  angled  triangle  EHG,  the  square  of  HE  is  to  the  square 
of  HG  as  the  segment  EF  is  to  the  segment  FG  (Prop.  XI. 
Cor.  3.),  or  as  M  is  to  N ;  hence  HP  :  HK^  :  :  M  :  N.  But 
HK=AB ;  therefore  the  square  described  upon  HI  is  to  the 
square  described  upon  AB  as  M  is  to  N.* 


PROBLEM  XII. 

Upon  a  given  line,  to  describe  a  polygon  similar  to  a  given 

polygon. 

Let  FG  be  the  given 
line,  and  AEDCB  the 
given  polygon. 

In  the  given  polygon, 
draw  the  diagonals  AC, 
AD;  at  the  point  F 
make  the  angle  GFH= 
BAC,  and  at  the  point 
G  the  angle  FGH=ABC  ;  the  lines  FH,  GH  will  cut  each 
other  in  H,  and  FGH  will  be  a  triangle  similar  to  ABC.  In 
the  same  manner  upon  FH,  homologous  to  AC,  describe  the 
triangle  FIH  similar  to  ADC  ;  and  upon  FI,  homologous  to  AD, 
describe  the  triangle  FIK  similar  to  ADE.  The  polygon 
FGHIK  will  be  similar  to  ABCDE,  as  required. 

For,  these  two  polygons  are  composed  of  the  same  number 
of  triangles,  which  are  similar  and  similarly  situated  (Prop. 
XXVLSch.). 


BOOK  IV. 


105 


PROBLEM  XIII. 


Tloo  similar  figures  being  given,  to  describe  a  similar  figure 
which  shall  be  equivalent  to  their  sum  or  their  difference. 

Let  A  and  B  be  two  homologous  sides  of  the  given  figures. 

Find  a  square  equivalent  to  the 
sum  or  to  the  difference  of  the 
squares  described  upon  A  and  B ; 
let  X  be  the  side  of  that  square  ; 
then  will  X  in  the  figure  required, 
be  the  side  which  is  homologous 
to  the  sides  A  and  B  in  the  given 
figures.  The  figure  itself  may  then 
be  constructed  on  X,  by  the  last  problem. 

For,  the  similar  figures  are  as  the  squares  of  their  homolo- 
gous sides ;  now  the  square  of  the  side  X  is  equivalent  to  the 
sum,  or  to  the  difference  of  the  squares  described  upon  the 
homologous  sides  A  and  B  ;  therefore  the  figure  described  upon 
the  side  X  is  equivalent  to  the  sum,  or  to  the  difference  of  the 
similar  figures  described  upon  the  sides  A  and  B. 


PROBLEM  XIV. 


To  describe  a  figure  similar  to  a  given  figure,  and  bearing  to  it 
the  given  ratio  of  M  to  N. 


Let  A  be  a  side  of  the  given  figure,  X 
the  homologous  side  of  the  figure  required. 
The  square  of  X  must  be  to  the  square  of 
A,  as  M  is  to  N  :  hence  X  will  be  found  by 
(Prob.  XL),  and  knowing  X,  the  rest  will  be 
accomplished  by  (Prob.  XIL). 


106 


GEOMETRY. 


PROBLEM  XV. 


To  construct  a  figure  similar  to  the  figure  P,  and  equivalent  to 
the  figure  Q. 

Find  M,  the  side  of  a  square 
equivalent  to  the  figure  P,  and 
N,  the  side  of  a  square  equiva- 
lent to  the  figure  Q.  Let  X  be 
a  fourth  proportional  to  the  three 
given  lines,  M,  N,  AB  ;  upon 
the  side  X,  homologous  to  AB, 
describe  a  figure  similar  to  the  figure  P ;  it  will  also  be  equiva- 
lent to  the  figure  Q. 

For,  calling  Y  the  figure  described  upon  the  side  X,  we  have 
P  ;  Y  :  :  AB2  :  X^ ;  but  by  construction,  AB  :  X  :  :  M  :  N, 
or  AB2  :  X^  :  :  M2  :  N2 ;  hence  P  :  Y  :  :  M'^  :  W,  But  by 
construction  also,  M-=P  and  N2=Q;  therefore  P  :  Y  :  :  P  : 
Q;  consequently  Y=Q;  hence  the  figure  Y  is  similar  to  the 
figure  P,  and  equivalent  to  the  figure  Q. 


PROBLEM  XVI. 


To  construct  a  rectangle  equivalent  to  a  given  square,  and  having 
the  sum  of  its  adjacent  sides  equal  to  a  given  line. 

Let  C  be  the  square,  and  AB  equal  to  the  sum  of  the  sides 
of  the  required  rectangle. 

Upon  AB  as  a  diame- 
ter, describe  a  semicir- 
cle ;  draw  the  line  DE 
parallel  to  the  diameter, 
at  a  distance  AD  from  it,       __ 

equal  to  the  side  of  the     A  1?]3 

given  square  C ;  from  the  point  E,  where  the  parallel  cutp  the 
circumference,  draw  EF  perpendicular  to  the  diameter ;  AF 
and  FB  will  be  the  sides  of  the  rectangle  required. 

For  their  sum  is  equal  to  AB  ;  and  their  rectangle  AF.FB  is 
equivalent  to  the  square  of  EF,  or  to  the  square  of  AD ;  hence 
that  rectangle  is  equivalent  to  the  given  square  C. 

Scholium,  To  render  the  problem  possible,  the  distance  AD 
must  not  exceed  the  radius ;  that  is,  the  side  of  the  square  C 
must  not  exceed  the  half  of  the  line  AB. 


BOOK  IV. 


107 


PROBLEM  XVII. 


To  construct  a  rectangle  that  shall  he  equivalent  to  a  given 
square,  and  the  difference  of  whose  adjacent  sides  shall  he 
equal  to  a  given  line. 

Suppose  C  equal  to  the  given  square,  and  AB  the  difference 
of  the  sides. 

Upon  the  given  line  AB  as  a  diame- 
ter, describe  a  semicircle  :  at  the  ex- 
tremity of  the  diameter  drav^^  the  tan- 
gent AD,  equal  to  the  side  of  the  square 
C  ;  through  the  point  D  and  the  centre 
O  draw  the  secant  DF ;  then  will  DE 
and  DF  be  the  adjacent  sides  of  the 
rectangle  required. 

For,  first,  the  difference  of  these  sides 
is  equal  to  the  diameter  EF  or  AB  ; 
secondly,  the  rectangle   DE,  DF,   is 
equal  to  AD^  (Prop.  XXX.) :  hence  that  rectangle  is  equivalent 
to  the  given  square  C. 


PROBLEM  XVIII. 


To  find  the  common  measure,  if  there  is  one,  between  the  diagonal 
and  the  side  of  a  square. 


Let  ABCG  be  any  square  what- 
ever, and  AC  its  diagonal. 

We  must  first  apply  CB  upon 
CA,  as  often  as  it  may  be  contained 
there.  For  this  purpose,  let  the 
semicircle  DBE  be  described,  from 
the  centre  C,  with  the  radius  CB. 
It  is  evident  that  CB  is  contained 
once  in  AC,  with  the  remainder 
AD ;  the  result  of  the  first  operation 
is  therefore  the  quotient  1,  with  the  remainder  AD,  which  lat- 
ter must  now  be  compared  with  BC,  or  its  equal  AB. 

We  might  here  take  AF=AD,  and  actually  apply  it  upon 
AB  ;  we  should  find  it  to  be  contained  twice  with  a  remain- 
der :  but  as  that  remainder,  and  those  which  succeed  it,  con- 


108 


GEOMETRY. 


tinue  diminishing,  and  would  soon 
elude  our  comparisons  by  their  mi- 
nuteness, this  would  be  but  an  imper- 
fect mechanical  method,  from  which 
no  conclusion  could  be  obtained  to 
determine  whether  the  lines  AC,  CB, 
have  or  have  not  a  common  measure. 
There  is  a  very  simple  way,  however, 
of  avoiding  these  decreasing  lines, 
and  obtaining  the  result,  by  operating 
only  upon  lines  which  remain  always  of  the  same  magnitude. 

The  angle  ABC  being  a  right  angle,  AB  is  a  tangent,  and 
4E  a  secant  drawn  from  the  same  point ;  so  that  AD  :  AB  :  : 
AB  :  AE  (Prop.  XXX.).  Hence  in  the  second  operation,  when 
AD  is  compared  with  AB,  the  ratio  of  AB  to  AE  may  be  taken 
instead  of  that  of  AD  to  AB  ;  now  AB,  or  its  equal  CD,  is  con- 
tained twice  in  AE,  with  the  remainder  AD  r  the  result  of  the 
second  operation  is  therefore  the  quotient  2  with  the  remain- 
der AD,  which  must  be  compared  with  AB. 

Thus  the  third  operation  again  consists  in  comparing  AD' 
with  AB,  and  may  be  reduced  in  the  same  manner  to  the  com- 
parison of  AB  or  its  equal  CD  with  AE  ;  from  which  there  will 
again  be  obtained  2  for  the  quotient,  and  AD  for  the  re- 
mainder. 

Hence,  it  is  evident  that  the  process  will  never  terminate  ; 
and  therefore  there  is  no  common  measure  between  the  diago- 
nal and  the  side  of  a  square  :  a  truth  which  was  already  known 
by  arithmetic,  since  these  two  lines  are  to  each  other  :  :  V2  :  1 
(Prop.  XI.  Cor.  4.),  but  which  acquires  a  greater  degree  of 
clearness  by  the  g-eometrical  investigation. 


BOOK  V.  100 


BOOK  V. 


REGULAR  POLYGONS,  AND  THE  MEASUREMENT  OF  THE 
CIRCLE. 

Definition, 

A  Polygon,  which  is  at  once  equilateral  and  equiangular,  is 
called  a  regular  polygon. 

Regular  polygons  may  have  any  number  of  sides :  the  equi- 
lateral triangle  is  one  of  three  sides  ;  the  square  is  one  of  four. 


PROPOSITION  I.  THEOREM. 

Two  regular  polygons  of  the  same  number  of  sides  are  similar 

figures. 

Suppose,  for  example, 
that  ABCDEF,  ahcdefi 
are  two  regular  hexa- 
gons. The  sum  of  all  the 
angles  is  the  same  in  both 
figures,being  in  each  equal 
to  eight  right  angles  (Book  I.  Prop.  XXVI.  Cor.  3.).  The  angle 
A  is  the  sixth  part  of  that  sum ;  so  is  the  angle  a  :  hence  the 
angles  A  and  a  are  equal ;  and  for  the  same  reason,  the  angles 
B  and  5,  the  angles  C  and  c,  &c.  are  equal. 

Again,  since  the  polygons  are  regular,  the  sides  AB,  BC,  CD, 
&c.  are  equal,  and  likewise  the  sides  ah,  he,  cd,  &c.  (Def )  ;  it  is 
plain  that  AB  :  ah  :  -.  BC  i  he  i  :  CD  :  cd,  &c. ;  hence  the 
two  figures  in  question  have  their  angles  equal,  and  their  ho- 
mologous sides  proportional ;  consequently  they  are  similar 
(Book  IV.  Def  1.). 

Cor.  The  perimeters  of  two  regular  polygons  of  the  same 
number  of  sides,  are  to  each  other  as  their  homologous  sides, 
and  their  surfaces  are  to  each  other  as  the  squares  of  those  sides 
(Book  IV.  Prop.  XXVIL). 

Scholium.  The  angle  of  a  regular  polygon,  like  the  angle  of 
an  equiangular  polygon,  is  determined  by  the  number  of  its 
Bides  (Book  I.  Prop.  XXVL). 

K 


no  GEOMETRY. 


PROPOSITION  II.    THEOREM. 


Any  regular  polygon  may  he  inscribed  in  a  circle^  and  circum- 
scribed about  one. 

Let  ABCDE  &c.  be  a  regular  poly- 
gon :  describe  a  circle  through  the  three 
points  A,  B,  C,  the  centre  being  O,  and 
OP  the  perpendicular  let  fall  from  it,  to 
the  middle  point  of  BC :  draw  AO  and 
OD. 

If  the  quadrilateral  OPCD  be  placed 
upon  the  quadrilateral  OPBA,  they  will  _ 

coincide  ;  for  the  side  OP  is  common  ;  IP 

the  angle  OPC=:OPB,  each  being  a  right  angle  ;  hence  the 
side  PC  will  apply  to  its  equal  PB,  and  the  point  C  will  fall  on 
B :  besides,  from  the  nature  of  the  polygon,  the  angle  PCD=: 
PBA ;  hence  CD  will  take  the  direction  BA ;  and  since  CD=: 
BA,  the  point  D  will  fall  on  A,  and  the  two  quadrilaterals  will 
entirely  coincide.  The  distance  OD  is  therefore  equal  to  AO  ; 
and  consequently  the  circle  which  passes  through  the  three 
points  A,  B,  C,  will  also  pass  through  the  point  D.  By  the 
same  mode  of  reasoning,  it  might  be  shown,  that  the  circle 
which  passes  through  the  three  points  B,  C,  D,  will  also  pass 
through  the  point  E  ;  and  so  of  all  the  rest :  hence  the  circle 
which  passes  through  the  points  A,  B,  C,  passes  also  through 
the  vertices  of  all  the  angles  in  the  polygon,  which  is  therefore 
inscribed  in  this  circle. 

Again,  in  reference  to  this  circle,  all  the  sides  AB,  BC,  CD, 
&c.  are  equal  chords ;  they  are  therefore  equally  distant  from 
the  centre  (Book  III.  Prop.  VIII.):  hence,  if  from  the  point  O 
with  the  distance  OP,  a  circle  be  described,  it  will  touch  the 
side  BC,  and  all  the  other  sides  of  the  polygon,  each  in  its  mid- 
dle point,  and  the  circle  will  be  inscribed  in  the  polygon,  or  the 
polygon  described  about  the  circle. 

Scholium  1.  The  point  O,  the  common  centre  of  the  in- 
scribed and  circumscribed  circles,  may  also  be  regarded  as  the 
centre  of  the  polygon  ;  and  upon  this  principle  the  anglo  AOB 
is  called  the  angle  at  the  centre,  being  formed  by  two  radii 
drawn  to  the  extremities  of  the  same  side  AB. 

Since  all  the  chords  AB,  BC,  CD,  &c.  are  equal,  all  the  an- 
gles at  the  centre  must  evidently  be  equal  likewise  ;  and  there- 
fore the  value  of  each  will  be  found  by  dividing  four  right  an- 
gles by  the  number  of  sides  of  the  polygon. 


BOOK  V. 


Ill 


Scholium  2.  To  inscribe  a  regu- 
lar polygon  of  a  certain  number  of 
sides  in  a  given  circle,  we  have  only 
to  divide  the  circumference  into  as 
many  equal  parts  as  the  polygon 
has  sides :  for  the  arcs  being  equal, 
the  chords  AB,  BC,  CD,  &c.  will 
also  be  equal ;  hence  likewise  the 
triangles  AOB,  BOC,  COD,  must 
be  equal,  because  the  sides  are  equal  each  to  each  ;  hence  all 
the  angles  ABC,  BCD,  CDE,  &c.  will  be  equal ;  hence  the 
figure  ABCDEH,  will  be  a  regular  polygon. 


PROPOSITION  III.     PROBLEM. 


To  inscribe  a  square  in  a  given  circle. 


Draw  two  diameters  AC,  BD,  cut- 
ting each  other  at  right  angles  ;  join 
their  extremities  A,  B,  C,  D :  the  figure 
ABCD  will  be  a  square.  For  the  an- 
gles AOB,  BOC,  &c.  being  equal,  the 
chords  AB,  BC,"  &c.  are  also  equal : 
and  the  angles  ABC,  BCD,  &c.  being 
in  semicircles,  are  right  angles. 


Scholium.  Since  the  triangle  BCO  is  right  angled  and  isos- 
celes, we  have  BC  :  BO  :  :  v/2  :  1  (Book  IV.  Prop.  XI. 
(Jor.  4.) ;  hence  the  side  of  the  inscribed  square  is  to  the  radius^ 
as  the  square  root  of  2,  is  to  unity. 


PROPOSITION  IV.      PROBLEM. 


In  a  given  circle,  to  inscribe  a  regular  hexagon  and  an  equilate- 


ral triangle. 


112 


GEOMETRY. 


Suppose  the  problem  solved, 
and  that  AB  is  a  side  of  the  in- 
scribed hexagon  ;  the  radii  AO, 
OB  being  drawn,  the  triangle 
AOB  will  be  equilateral. 

For,  the  angle  AOB  is  the  sixth 
part  of  four  right  angles  ;  there- 
fore, taking  the  right  angle  for 
unity,  we  shall  have  AOB=|  — 
f :  and  the  two  other  angles 
ABO,  BAO,  of  the  same  trian- 
gle, are  together  equal  to  2 — | 
=f  ;  and  being  mutually  equal,  _ 

each  of  them  must  be  equal  to  | ;  hence  the  triangle  ABO  W 
equilateral ;  therefore  the  side  of  the  inscribed  hexagon  is  equal 
to  the  radius. 

Hence  to  inscribe  a  regular  hexagon  in  a  given  circle,  the 
radius  must  be  applied  six  times  to  the  circumference  ;  which 
will  bring  us  round  to  the  point  of  beginning. 

And  the  hexagon  ABCDEF  being  inscribed,  the  equilateral  < 
triangle  ACE  may  be  formed  by  joining  the  vertices  of  the 
alternate  angles. 

Scholium,  The  figure  ABCO  is  a  parallelogram  and  even 
a  rhombus,  since  AB=BC=CO=AO  ;  hence' the  sum  of  the 
squares  of  the  diagonals  AC^+BO-  is  equivalent  to  the  sum  of 
the  squares  of  the  sides,  that  is,  to  4AB^  or  4B0^  (Book  IV. 
Prop  XiV.  Cor.) :  and  taking  away  BO^  from  both,  there  will 
remain  AC2=3BO^  hence  AC^  :  BO^  :  :  3  :  l,orAC  :  BO 
:  :  \/3  :  1  ;  hence  the  side  of  the  inscribed  equilateral  triangle 
is  to  the  radius  as  the  square  root  of  three  is  to  unitu* 


PROPOSITION  V.     PROBLEM. 


In  a  given  circle^  to  inscribe  a  regular  decagon;  then  apentagony 
and  also  a  regular  polygon  of  fifteen  sides. 


BOOK  V. 


113 


Divide  the  radius  AO  in 
extreme  and  mean  ratio  at 
the  point  M  (Book  IV.  Prob. 
IV.) ;  take  the  chord  AB  equal 
to  OM  the  greater  segment ; 
AB  will  be  the  side  of  the 
regular  decagon,  and  will  re- 
quire to  be  applied  ten  times 
to  the  circumference. 

For,  drawing  MB,  we  have 
by  construction,  AO  :  OM 
:  :  OM  :  AM  ;  or,  since  AB 
C.OM,  AO  :  AB  :  :  AB  : 
AM  ;  since  the  triangles  ABO,  AMB,  have  a  common  angle  A, 
included  between  proportional  sides,  they  are  similar  (Book 
IV.  Prop.  XX.).  Now  the  U'iangle  OAB  being  isosceles,  AMB 
must  be  isosceles  also,  and  AB=BM  ;  but  AB=:OM  ;  hence 
also  MB  =  OM  ;  hence  the  triangle  BMO  is  isosceles. 

Again,  the  angle  AMB  being  exterior  to  the  isosceles  trian- 
gle BMO,  is  double  of  the  interior  angle  O  (Book  I.  Prop. 
XXV.  Cor.  6,)  :  but  the  angle  AMBzrrMAB  ;  hence  the  trian- 
gle OAB  is  such,  that  each  of  the  angles  OAB  or  OBA,  at  its 
base,  is  double  of  O,  the  angle  at  its  vertex  ;  hence  the  three 
angles  of  the  triangle  are  together  equal  to  five  times  the  angle 
O,  which  consequently  is  the  fifth  part  of  the  two  right  angles, 
or  the  tenth  part  of  four  ;  hence  the  arc  AB  is  the  tenth  part 
of  the  circumference,  and  the  chord  AB  is  the  side  of  the  reg- 
ular decagon. 

2d.  By  joining  the  alternate  corners  of  the  regular  decagon, 
the  pentagon  ACEGI  will  be  formed,  also  regular. 

3d.  AB  being  still  the  side  of  the  decagon,  let  AL  be  the 
side  of  a  hexagon  ;  the  arc  BL  will  then,  with  reference  to 
the  whole  circumference,  be  } — j\,  or  yV  ;  hence  the  chord  BL 
will  be  the  side  of  the  regular  polygon  of  fifteen  sides,  or  pente- 
decagon.  It  is  evident  also,  that  the  arc  CL  is  the  third  of  CB. 

Scholium.  Any  regular  polygon  being  inscribed,  if  the  arcs 
subtended  by  its  sides  be  severally  bisected,  the  chords  of  those 
semi-arcs  will  form  a  new  regular  polygon  of  double  the  num- 
ber of  sides  :  thus  it  is  plain,  that  the  square  will  enable  us  to  in- 
scribe successively  regular  polygons  of  8, 16, 32,  &c.  sides.  And 
m  like  manner,  by  means  of  the  hexagon,  regular  polygons  of 
12,  24,  48,  &c.  sides  may  be  inscribed  ;  by  means  of  the  deca- 
gon, polygons  of  20, 40, 80,  &c.  sides  ;  by  means  of  the  pente- 
decagon,  polygons  of  30,  60,  120,  &c.  sides. 

It  is  further  evident,  that  any  of  the  inscribed  polygons  will 
be  less  than  the  inscribed  polygon  of  double  the  number  of 
sides,  since  a  part  is  less  than  the  whole. 

K* 


114 


GEOMETRY. 


PROPOSITION  VI.    PROBLEM. 


A  regular  inscribed  poly gon  being  gimn^  to  circumscribe  a  sim 
ilar  polygon  about  the  same  circle. 

Let  CBAFED  be  a  regular  polygon. 

At  T,  the  middle  point 
of  the  arc  AB,  apply  the 
tangent  GH,  which  will 
be  parallel  to  AB  (Book 
III.  Prop.  X.)-;  do  the 
same  at  the  middle  point 
of  each  of  the  arcs  BC, 
CD,  &c. ;  these  tangents, 
by  their  intersections, 
will  form  the  regular 
circumscribed  polygon 
GHIK  &c.  similar  to 
the  one  inscribed.  _^ 

ik:       Q        ii 

Since  T  is  the  middle  point  of  the  arc  BTA,  and  N  the  mid- 
dle point  of  the  equal  arc  BNC,  it  follows,  that  BT=:BN  ;  or 
that  the  vertex  B  of  the  inscribed  polygon,  is  at  the  middle 
point  of  the  arc  NBT.  Draw  OH.  The  line  OH  will  pass 
through  the  point  B. 

For,  the  right  angled  triangles  OTH,  OHN,  having  the  com- 
mon hypothenuse  OH,  and  the  side  OT=:ON,  must  be  equal 
(Book  I.  Prop.  XVH.),  and  consequently  the  angle  TOHr:r 
HON,  wherefore  the  line  OH  passes  through  the  middle  point 
B  of  the  arc  TN.  For  a  like  reason,  the  point  1  is  in  the  pro- 
longation of  OC  ;  and  so  with  the  rest. 

But,  since  GH  is  parallel  to  AB,  and  HI  to  BC,  the  angle 
GHI=ABC  (Book  I.  Prop.  XXIV.) ;  in  like  manner  HIKrr 
BCD  ;  and  so  with  all  the  rest :  hence  the  angles  of  the  cir- 
cumscribed polygon  are  equal  to  those  of  the  inscribed  one. 
And  further,  by  reason  of  these  same  parallels,  we  have  GH  : 
AB  :  :  OH  :  OB,  and  HI  :  BC  :  :  OH  :  OB  ;  thefefore  GH  : 
AB  :  :  HI  :  BC.  But  AB=BC,  therefore  GHrrHI.  For  the 
same  reason,  HI =IK,  &c.;  hence  the  sides  of  the  circum- 
scribed polygon  are  all  equal ;  hence  this  polygon  is  regular, 
and  similar  to  the  inscribed  one. 

Ccr.  1.  Reciprocally,  if  the  circumscribed  polygon  GHIK 
&c.  were  given,  and  the  inscribed  one  ABC  &c.  were  re- 
quired to  be  deduced  from  it,  it  would  only  be  necessary  to 


BOOR  V. 


115 


draw  from  the  angles  G,  H,  I,  &c 
lines  OG,  OH,  &c.  meeting  the 
A,  B,  C,  &c. ;  then  to  join 
AB,  BC,  &c. ;  this  would  form 
easier  solution  of  this  problem 
points  of  contact  T,  N,  P,  <fec. 
which  likewise  would  form  an 
the  circumscribed  one. 


.  of  the  given  polygon,  straight 
circumference  in  the  points 

those  points  by  the  chords 
the  inscribed  polygon.     An 

would  be  simply  to  join  the 

by  the  chords  TN,  NP,  &c. 

inscribed  polygon  similar  to 


Cor,  2.  Hence  we  may  circumscribe  about  a  circle  any 
regular  polygon,  which  can  be  inscribed  within  it,  and  con- 
versely. 

Cor.  3.  It  is  plain  that  NH  +  HT=rHT  +  TG=HG,  one  of 
the  equal  sides  of  the  polygon. 

PROPOSITION  VII.    PROBLEM. 


A  circle  and  regular  circumscribed  polygon  being  given,  it  is 
required  to  circumscribe  the  circle  by  another  regular  polygon 
having  double  the  number  of  sides. 

Let  the  circle  whose  centre  is  P,  be  circumscribed  by  the 
square  CDEG ;  it  is  required  to  find  a  regular  circumscribed 
octagon. 

Bisect  the  arcs  AH,  HB,  BF, 
FA,  and  through  the  middle 
points  c,  d,  a,  b,  draw  tangenls  to 
the  circle,  and  produce  them  till 
they  meet  the  sides  of  the  square : 
then  will  the  figure  ApKdB  &c. 
be  a  regular  octagon. 

For,  having  drawn  Vd,  Va,  let 
the  quadrilateral  P^^B,  be  ap- 
plied to  the  quadrilateral  PB/a, 
so  that  PB  shall  fall  on  PB. 
Then,  since  the  angle  dVB  is 
equai  to  the  angle  BPa,  each  being  half  a  right  angle,  the  line 
Vd  will  fall  on  its  equal  Va,  and  the  point  d  on  the  point  a.  But 
the  angles  Vdg,  Vaf,  are  right  angles  (Book  HI.  Prop.  IX.)  ; 
hence  the  line  dg  will  take  the  direction  af.  The  angles  PB^, 
PB/,  are  also  right  angles  ;  hence  B^  will  take  the  direction 
Bf ;  therefore,  the  two  quadrilaterals  will  coincide,  and  the 
point  ^  will  fall  at/;  hence,  B^=Bf,  c?^=«/,  and  the  angle 
dgB  =  Bfa.  By  applying  in  a  similar  manner,  the  quadrilate- 
rals BBfa,  VFha,  it  may  be  shown,  that  af^ah,  fB=Fh,  and 
the  angle  Bfa—ahF.     But  since  the  two* tangents /z,  /B,  are 


116  GEOMETRY. 

equal  (Book  III.  Prob.  XIV.  Sch.),  it  follows  that  fh,  which  is 
twice /a,  is  equal  to^,  which  is  twice /B. 

In  a  similar  manner  it  may  be  shown  that  /(/r=/ii,  and  the 
angle  Yit=Yha,  or  that  any  two  sides  or  any  two  angles  of  the 
octagon  are  equal :  hence  the  octagon  is  a  regular  polygon  (Def.). 
The  construction  which  has  been  made  in  the  case  of  the  square 
and  the  octagon,  is  equally  applicable  to  other  polygons. 

Cor  It  is  evidentthat  the  circumscribed  square  is  greater  than 
the  circumscribed  octagon  by  the  four  triangles,  Cnp,  kDgf 
hEf,  Git ;  and  if  a  regular  polygon  of  sixteen  sides  be  circum- 
scribed about  the  circle,  we  may  prove  in  a  similar  way,  that 
the  figure  having  the  greatest  number  of  sides  will  be  the  least ; 
and  the  same  may  be  shown,  whatever  be  the  number  of  sides 
of  the  polygons  :  hence,  in  general,  any  circumscribed  regular 
polygon,  will  he  greater  than  a  circumscribed  regular  polygon 
having  double  the  number  of  sides. 


PROPOSITION  VIII.    THEOREM. 

Two  regular  polygons,  of  the  same  number  of  sides,  can  always 
be  formed,  the  one  circumscribed  about  a  circle,  the  other  in- 
scribed in  it,  which  shall  differ  from  each  other  by  less  than 
any  assignable  surface. 

Let  Q  be  the  side  of  a  square 
less  than  the  given  surface. 
Bisect  AC,  a  fourth  pare  of 
the  circumference,  and  then  bi 
sect  the  half  of  this  fourth,  and 
proceed  in  this  manner,  always 
bisecting  one  of  the  arcs  formed  ^ 
by  the  last  bisection,  until  an 
arc  is  found  whose  chord  AB  is  - 
less  than  Q.  As  this  arc  will 
be  an  exact  part  of  the  circum- 
ference, if  we  apply  chords  AB, 
BC,  CI),  &c.  each  equal  to  AB,  the  last  will  terminate  at  A, 
and  there  will  be  formeji  a  regular  polygon  ABCDE  &c.  in 
the  circle. 

Next,  describe  about  the  circle  a  similar  polygon  abcde  (fee. 
(Prop.  VI.) :  the  difference  of  these  two  polygons  will  be  less 
than  the  square  of  Q. 

For,  from  the  points  a  and  b,  draw  the  lines  aO,  bO,  to  the 
centre  O  :  they  will  pass  through  the  points  A  and  B,  as  was 


BOOK  V.  117 

shown  in  Prop.  VI.  Draw  also  OK  to  the  point  of  contact 
K  :  it  will  bisect  AB  in  I,  and  be  perpendicular  to  it  (Book  III. 
Prop.  VI.  Sch.).     Produce  AO  to  E,  and  draw  BE. 

Let  P  represent  the  circumscribed  polygon,  and  p  the  in- 
scribed polygon  :  then,  since  the  triangles  aOb,  AOB,  are  like 
parts  of  P  and  p,  we  shall  have 

aOb  -i  AOB  :  :  P  :  p  (Book  II.  Prop.  XI.)  : 
But  the  triangles  being  similar, 

aOb  :  AOB  :  :  Oa''  :  0A«,  or  OK^ 
Hence,         P  :  p  :   :   Oa^  :   OK^ 

Again,  since  the  triangles  0«K,  EAB  are  similar,  havmg 
their  sides  respectively  parallel, 

Oa'^  :  OK^  :  :  AE^  r  EB\  hence, 
F  :  p  :  :  AE^  :  EB^  or  by  division, 

P  :  P-p  :  :  AE=^  :  AE^^-EB^  or  AB^. 
But  P  is  less  than  the  square  described  on  the  diameter  AE 
(Prop.  VII.  Cor.);  therefore  F—p  is  less  than  the  square  de- 
scribed on  AB ;  that  is,  less  than  the  given  square  on  Q :  hence 
the  difference  between  the  circumscribed  and  inscribed  poly- 
gons may  always  be  made  less  than  a  given  surface. 

Cor.  1.  A  circumscribed  regular  polygon,  having  a  given 
number  of  sides,  is  greater  than  the  circle,  because  the  circle 
makes  up  but  a  part  of  the  polygon  :  and  for  a  like  reason,  the 
inscribed  polygon  ie  less  than  the  circle.  But  by  increasing 
the  number  of  sides  of  the  circumscribed  polygon,  the  polygon 
is  diminished  (Prop.  VII,  Cor.),  and  therefore  approaches  to 
an  equality  with  the  circle  ;  and  as  the  number  of  sides  of 
the  inscribed  polygon  is  increased,  the  polygon  is  increased 
(Prop.  V.  Sch.),  and  therefore  approaches  to  an  equality  with 
the  circle. 

Now,  if  the  number  of  sides  of  the  polygons  he  indefinitely  in- 
creased, the  length  of  each  side  will  be  indefinitely  small,  and  the 
polygons  will  ultimately  become  equal  to  each  other,  and  equal 
also  to  the  circle. 

For,  if  they  are  not  ultimately  equal,  let  D  represent  their 
smallest  difference. 

Now,  it  has  been  proved  in  the  proposition,  that  the  differ- 
ence between  the  circumscribed  and  inscribed  polygons,  can 
be  made  less  than  any  assignable  quantity  :  that  is,  less  than 
D  :  hence  the  difference  between  tlie  oolygons  is  equal  to  D, 
and  less  than  D  at  the  same  time,  which  is  absurd  :  therefore, 
the  polygons  are  ultimately  equal.  But  when  they  are  equal 
to  each  other,  each  must  also  be  equal  to  the  circle,  since  the 
circumscribed  polygon  cannot  fair  within  the  circle,  nor  the 
inscribed  polygon  without  it. 


118 


GEOMETRY. 


Cor.  2.  Since  the  circumscribed  polygon  has  the  same  num- 
ber of  sides  as  the  corresponding  inscribed  polygon,  and  since 
the  two  polygons  are  regular,  they  will  be  similar  (Prop,  I.)  ; 
and  therefore  when  they  become  equal,  they  will  exactly  coin- 
cide, and  have  a  common  perimeter.  But  as  the  sides  of  the 
circumscribed  polygon  cannot  fall  within  the  circle,  nor  the 
sides  of  the  inscribed  polygon  without  it,  it  follows  that  the 
perimeters  of  the  polygons  will  unite  on  the  circumference  of  the 
circle,  and  become  equal  to  it. 

Cor.  3.  When  the  number  of  sides  of  the  inscribed  polygon 
is  indefinitely  increased,  and  the  polygon  coincides  with  the 
circle,  the  line  01,  drawn  from  the  centre  O,  perpendicular  to 
the  side  of  the  polygon,  will  become  a  radius  of  the  circle,  and 
any  portion  of  the  polygon,  as  ABCO,  will  become  the  sector 
OAKBC,  and  the  part  of  the  perimeter  AB  +  BC,  will  become 
thearcAKBC. 


PROPOSITION  IX.    THEOREM. 


Tile  area  of  a  regular  polygon  is  equal  to  its  perimeter,  multi- 
plied by  half  the  radius  of  the  inscribed  circle. 

Let  there  be  the  regular  polygon 
GHIK,  and  ON,  OT,  radii  of  the  in- 
scribed circle.  The  triangle  GOH 
will  be  measured  by  GH  x  ^OT ;  the 
triangle  OHI,  by  HIxiON:  but 
ON^^OT;  hence  the  two  triangles 
taken  together  will  be  measured  by 
(GH  +  HI)xiOT.  And,  by  con- 
tinuing the  same  operation  for  the 
other  triangles,  it  will  appear  that 
the  sum  of  them  all,  or  the  whole 
polygon,  is  measured  by  the  sum  of  the  bases  GH,  HI,  &c. 
or  the  perimeter  of  the  polygon,  multiplied  into  ^OT,  or  half 
the  radius  of  the  inscribed  circle. 

Scholium.  The  radius  OT  of  the  inscribed  circle  is  nothing 
else  than  the  perpendicular  let  fall  from  the  centre  on  one  of 
the  sides :  it  is  sometimes  named  the  apothem  of  the  polygon. 


BOOK  V. 


no 


PROPOSITION  X.    THEOREM. 


The  perimeters  of  two  regular  polygons,  having  the  same  num- 
ber of  sides,  are  to  each  other  as  the  radii  of  the  circumscribed 
circles,  and  also,  as  the  radii  of  the  inscribed  circles ;  and  their 
areas  are  to  each  other  as  the  squares  of  those  radii. 

Let  AB  be  the  side  of  the  one  poly- 
gon, O  the  centre,  and  consequently 
OA  the  radius  of  the  circumscribed 
circle,  and  OD,  perpendicular  to  AB, 
the  radius  of  the  inscribed  circle ;  let 
ah,  in  like  manner,  be  a  side  of  the 
other  polygon,  o  its  centre,  oa  and  od 
the  radii  of  the  circumscribed  and  the 
inscribed  circles.  The  perimeters  of 
the  two  polygons  are  to  each  other  as  the  sides  AB  and  ah 
(Book  IV.  Prop.  XXVII.) :  but  the  angles  A  and  a  are  equal, 
being  each  half  of  the  angle  of  the  polygon  ;  so  also  are  the 
angles  B  and  b ;  hence  the  triangles  ABO,  abo  are  similar,  as 
are  likewise  the  right  angled  triangles  ADO,  ado ;  hence 
AB  :  ab  :  :  AO  :  ao  :  :  DO  :  do  ;  hence  the  perimeters  of  the 
polygons  are  to  each  other  as  the  radii  AO,  ao  of  the  circum- 
scribed circles,  and  also,  as  the  radii  DO,  do  of  the  inscribed 
circles. 

The  surfaces  of  these  polygons  are  to  each  other  as  the 
squares  of  the  homologous  sides  AB,  ab  ;  they  are  therefore 
likewise  to  each  other  as  the  squares  of  AO,ao,the  radii  of  the 
circumscribed  circles,  or  as  the  squares  of  OD,  oc?,the  radii  of 
the  inscribed  circles. 


PROPOSITION  XI.    THEOREM. 


The  circumferences  of  circles  are  to  each  other  as  their  radii, 
and  the  areas  are  to  each  other  as  the  squares  of  their  radii. 


120 


GEOMETRY. 


Let  us  designate  the  circumference  of  the  circle  whose  radius 
is  CA  by  arc.  CA ;  and  its  area,  by  area  CA :  it  is  then  to  be 
shown  that 

circ,  CA  :  circ,  OB  :  :  CA  :  OB,  and  that 
area  CA  :  area  OB  :  :  CA^  :  OB^ 


Inscribe  within  the  circles  two  regular  polygons  of  the  same 
number  of  sides.  Then,  whatever  be  the  number  of  sides, 
their  perimeters  will  be  to  each  other  as  the  radii  CA  and  OB 
(Prop.  X.).  Now,  if  the  arcs  subtending  the  sides  of  the  poly- 
gons be  co/itinually  bisected,  until  the  number  of  sides  of  the 
polygons  shall  be  indefinitely  increased,  the  perimeters  of  the 
polygons  will  become  equal  to  the  circumferences  of  the  cir- 
cumscribed circles  (Prop.  VIII.  Cor.  2.),  and  we  shall  have 
circ,  CA  :  circ,  OB  :  :  CA  :  OB. 

Again,  the  areas  of  the  inscribed  polygons  are  to  each  other 
as  CA^  to  OB^  (Prop.  X.).  But  when  the  number  of  sides  of 
the  polygons  is  indefinitely  increased,  the  areas  of  the  polygons 
become  equal  to  the  areas  of  the  circles,  each  to  each,  (Prop. 
VIII.  Cor.  1.)  ;  hence  we  shall  have 

area  CA  :  area  OB  :  :  CA^  :  OB^. 


V 


:e 


Cor,  The  similar  arcs  AB, 
DE  are  to  each  other  as  their 
radii  AC,  DO  ;  and  the  similar 
sectors  ACB,  DOE,  are  to  each 
other  as  the  squares  of  their 
radii. 

For,  since  the  arcs  are  simi-  ^ ' 

lar,  the  angle  C  is  equal  to  the  angle  O  (Book  IV.  Def.  3.)  ; 
but  C  is  to  four  right  angles,  as  the  arc  AB  is  to  the  whole  cir- 
cumference described  with  the  radius  AC  (Book  III.  Prop. 
XVII.) ;  and  O  is  to  the  four  right  angles,  as  the  arc  DE  is  to 
the  circumference  described  with  the  radius  OD  :  hence  the 
arcs  AB,  DE,  are  to  each  other  as  the  circumferences  of  which 


BOOK  V.  121 

they  form  part :  but  these  circumferences  are  to  each  other  as 
their  radii  AC,  DO  ;  hence 

arc  AB  :  arc  DE  :  :  AC  :  DO. 
For  a  Hke  reason,  the  sectors  ACB,  DOE  are  to  each  other 
as  the  whole  circles ;  which  again  are  as  the  squares  of  their 
radii ;  therefore 

sect,  ACB  :  sect.  DOE  ;  :  AC^  :  D0\ 


PROPOSITION  XII.     THEOREM. 

The  area  of  a  circle  is  equal  to  the  product  of  its  circumference  by 
half  the  radius. 

Let  ACDE  be  a  circle  whose 
centre  is  O  and  radius  OA  :  then 
will 

area  OA=^OAx  arc.  OA. 

For,  inscribe  in  the  circle  any  ^ 
regular  polygon,  and  draw  OF 
perpendicular  to  one  of  its  sides. 
Then  the  area  of  the  polygon 
will  be  equal  to  ^OF,  multiplied 
by  the  perimeter  (Prop.  IX.). 
Now,  let  the  number  of  sides  of  the  polygon  be  indefinitely 
increased  by  continually  bisecting  the  arcs  which  subtend  the 
sides :  the  perimeter  will  then  become  equal  to  the  circumfe- 
rence of  the  circle,  the  perpendicular  OF  will  become  equal  to 
OA,  and  the  area  of  the  polygon  to  the  area  of  the  circle 
(Prop.  VIIL  Cor.  1.  &  3.).  But  the  expression  for  the  area 
will  then  become 

area  OA=^OA  x  circ,  OA  : 
consequently,  the  area  of  a  circle  is  equal  to  the  product  of 
half  the  radius  into  the  circumference. 

Cor.  1.  The  area  of  a  sector  is  equal 
to  the  arc  of  that  sector  multiplied  by  half 
its  radius. 

For,  the  sector  ACE  is  to  the  whole 
circle  as  the  arc  AMB  is  to  the  whole 
circumference  ABD  (Book  III.  Prop. 
XVII.  Sch.  2.),  or  as  AMBx^AC  is  to 
ABDx^AC.  But  the  whole  circle  is 
equal  to  ABD  x  ^AC  ;  hence  the  sector 
ACB  is  measured  by  AMB  x  i  AC. 

L 


Iii2  GEOMETRY. 

Cor.  2.  Let  the  circumference  of  the 
.  circle  whose  diameter  is  unity,  be  denoted 
by  n:  then,  because  circumferences  are 
to  each  other  as  their  radii  or  diameters, 
we  shall  have  the  diameter  I  to  its  cir- 
cumference TT,  as  the  diameter  2CA  is 
to  the  circumference  whose  radius  is  CA, 
that  is,  1  :  7t  :  :  2CA  :  arc.  CA,  tliere- 
fore  circ.  CA=7r  x  2CA.  Multiply  both 
terms  by  iCA  ;  we  have  ]CJA  x  circ.  CA 
=  7tx  CA^  or  area  CA=7i  x  CA^ :  hence- the  area  of  a  circle  is 
equal  to  the  product  of  the  square  of  its  radius  by  the  constant 
number  ^,  which  represents  the  circumference  whose  diameter 
is  1,  or  the  ratio  of  the  circumference  to  the  diameter. 

In  like  manner,  the  area  of  the  circle,  w-hose  radius  is  OB, 
will  be  equal  to  ^r  x  OB'^ ;  but  tt  x  CA^  :  n  x  OB^  :  :  CA^  :  OB^ ; 
hence  the  areas  of  circles  are  to  each  other  as  the  squares  of 
their  radii,  which  agrees  with  the  preceding  theorem. 

Scholium.  We  have  already  observed,  that  the  problem  of 
the  quadrature  of  the  circle  consists  in  finding  a  square  equal 
in  surface  to  a  circle,  the  radius  cf  which  is  known.  Now  it 
has  just  been  proved-  that  a  circle  is  equivalent  to  the  rectangle 
contained  by  its  circumference  and  half  its  radius  ;  and  this 
rectangle  may  be  changed  into  a  square,  by  finding  a  mean 
proportional  between  its  length  and  its  breadth  (Book  IV. 
Frob.  III.).  To  square  the  circle,  therefore,  is  to  find  the  cir- 
cumference when  the  radius  is  given  ;  and  for  effecting  this,  it 
is  enough  to  know  the  ratio  of  tTie  circumference  to  its  radius, 
or  its  diameter. 

Hitherto  the  ratio  in  question  has  never  been  determined 
except  approximatively  ;  but  the  approximation  has  been  car- 
ried so  far,  that  a  knowledge  of  the  exact  ratio  would  afford 
no  real  advantage  whatever  beyond  that  of  the  approximate 
ratio.  Accordingly,  this  problem,  which  engaged  geometers 
so  deeply,  when  their  methods  of  approximation  were  less  per- 
fect, is  now  degraded  to  the  rank  of  those  idle  questions,  with 
w^hich  no  one  possessing  the  slightest  tincture  of  geometrical 
science  will  occupy  any  portion  of  his  time. 

Archimedes  showed  that  the  ratio  of  the  circumference  to 
the  diameter  is  included  between  3|^  and  3|f  ;  hence  3^  or 
\2  affords  at  once  a  pretty  accurate  approximation  to  the  num- 
ber above  designated  by  ?? ;  and  the  simplicity  of  this  first  ap- 
proximation has  brought  it  into  very  general  use.  Metius, 
for  the  same  number,  found  the  much  more  accurate  value  ?f  |. 
At  last  the  value  of  7r,-developed  to  a  certain  order  of  decimals, 
wasfound  by  other  calculators  to  be  3.1415926535897932,  6ic.i 


BOOK  V.  123 

and  some  have  had  patience  enough  to  continue  these  decimals 
to  the  hundred  and  twenty-seventh,  or  even  to  the  hundred 
and  fortieth  place.  Such  an  approximation  is  evidently  equi- 
valent to  perfect  correctness  :  the  root  of  an  imperfect  power 
is  in  no  case  more  accurately  known. 

Tiie  following  problem  will  exhibit  one  of  the  simplest  ele- 
nientary  methods  of  obtaining  those  approximations. 

PROPOSITION  XIII.    PROBLEM. 

The  surface  of  a  regular  inscribed  poly gon^  and  that  of  a  simi- 
lar polygon  circumscribed,  being  given;  to  find  the  surfaces  of 
the  regular  inscribed  and  circumscribed  polygons  having 
double  the  number  of  sides. 

Let  AB  be  a  side  of  the  given 
inscribed  polygon; EF, parallel  to 
AB,  a  side  of  the  circumscribed 
polygon  ;  C  the  centre  of  the  cir- 
cle. If  the  chord  AM  and  the 
tangents  AP,  BQ,  be  drawn,  AM 
will  be  a  side  of  the  inscribed 
polygon,  having  twice  the  num- 
ber of  sides;  and  AP+PM=  2PM 
or  PQ,  will  be  a  side  of  the  simi- 
lar circumscribed  polygon  (Prop.  - 
VI.  Cor.  3.).  Now,  as  the  same  ^ 
construction  will  take  place  at  each  of  the  angles  equal  to 
ACM,  it  will  be  sufficient  to  consider  ACM  by  itself,  the  tri- 
angles connected  with  it  being  evidently  to  each  other  as  the 
whole  polygons  of  which  they  form  part.  Let  A,  then,  be 
the  surface  of  the  inscribed  polygon  whose  side  is  AB,  B  that 
of  the  similar  circumscribed  polygon  ;  A'  the  surface  of  the 
polygon  whose  side  is  AM,  B'  that  of  the  similar  circumscribed 
polygon :  A  and  B  are  given  ;  we  have  to  find  A'  and  B'. 

First.  The  triangles  ACD,  ACM,  having  the  common  ver- 
tex A,  are  to  each  other  as  their  bases  CD,  CM  ;  they  are  like- 
wise to  each  other  as  the  polygons  A  and  A',  of  which  they 
form  part :  hence  A  :  A'  :  :  CD  :  CM.  Again,  the  triangles 
CAM,  CME,  having  the  common  vertex  M,  are  to  each  other 
as  their  bases  CA,  CE  ;  they  are  likewise  to  each  other  as  the 
polygons  A'  and  B  of  which  they  form  part ;  hence  A'  :  B  :  : 
CA  :  CE.  But  since  AD  and  ME  are  parallel,  we  have 
CD  :  CM  :  :  CA  :  CE;  hence  A  :  A'  :  :  A'  :  B  ;  hence  the 
polygon  xV,  one  of  those  required,  is  a  mean  proportional  between 
the  two  given  polygons  A  and  B  and  consequently  A' =  V  A  x  B. 


121  GEOMETRY. 

Secondly-  The  altitude  CM  be- 
ing common,  the  triangle  CPM  is 
to  Jthe  triangle  CPE  as  PM  is  to 
PE  ;  but  since  CP  bisects  the  an- 
gle MCE,  we  have  PM  :  PE  :  : 
CM  :  CE  (Book  IV.  Prop. 
XYII.)::CD  :  CA  :  :  A  :  A' : 
hence  CPM  :  CPE  :  :  A  :  A' ; 
and  consequently  CPM  :  CPM  + 
CPEorCME::A:A  +  A'.  But 
CMPA,  or  2CMP,  and  CME  are 
to  each  other  as  the  polygons  B'  ^ 

and  B,  of  which  they  form  part :  hence  B'  :  B  :  :  2A  :  A  +  A'. 
Now  A'  has  been  already  determined  ;  this  new  proportion  will 

serve  for  determining  B',  and  give  us  B'—        -  ^;  and  thus  by 

A+  A' 
means  of  the  polygons  A  and  B  it  is  easy  to  find  the  polygons 
A'  and  B',  which  shall  have  double  the  number  of  sides. 


PROPOSITION  XIV.    PROBLEM. 

To  find  the  approximate  ratio  of  the  circumference  to  the 
diameter. 

Let  the  radius  of  the  circle  be  1  ;  the  side  of  the  inscribed 
square  will  be  \/2  (Prop.  III.  Sch.),  that  of  the  circumscribed 
square  will  be  equal  to  the  diameter  2  ;  hence  the  surface  of 
the  inscribed  square  is  2,  and  that  of  the  circumscribed  square 
is  4.  Let  us  therefore  put  A:r=2,  and  B=4  ;  by  the  last  pro- 
position we  shall  find  the  inscribed  octagonA' r=  V8=2.8284271, 

1  r* 
and  the  circumscribed  octagon  B'=^-—t^= 3.3 137085.    The 

mscribed  and  the  circumscribed  octagons  being  thus  deter- 
mined, we  shall  easily,  by  means  of  them,  determine  the  poly- 
gons having  twice  the  number  of  sides.  We  have  only  in  this 
case  to  put  A=:2.8284271,  B  =  3.3137085  ;  we  shall  find  A'  = 

2A  B 

N/A.B  =  3.0614674,and  B'=j— ^,  =  3.1825979.  These  poly- 
gons of  16  sides  will  in  their  turn  enable  us  to  find  the  polygons 
of  32  ;  and  the  process  may  be  continued,  till  there  remains 
no  longer  any  difference  between  the  inscribed  and  the  cir- 
cumscribed polygon,  at  least  so  far  as  that  place  of  decimals 
where  the  computation  stops,  and  so  far  as  the  seventh  place, 
in  this  example.     Being  arrived  at  this  point,  we  shall  infer 


BOOK  V.  12? 

that  the  last  result  expresses  the  area  of  the  circle,  which, 
since  it  must  always  lie  between  the  inscribed  and  the  circum- 
scribed polygon,  and  since  those  polygons  agree  as  far  as  a 
certain  place  of  decimals,  must  also  agree  with  both  as  far  as 
the  same  place. 

We  have  subjoined  the  computation  of  those  polygons,  car- 
r^'ed  on  till  they  agree  as  far  as  the  seventh  place  of  decimals. 

Number  of  sides  Inscribed  polygon.  Circumscribed  polygon. 

4 2.0000000  ....  4.0000000 

8 2.8284271  ....  3.3187085 

IG 3.0G14674  ....  3.1825979 

32 3.1214451  ....  3.1517249 

64  ....     .  3.1365485  ....  3.1441184 

128 3.1403311  ....  3.1422236 

256 3.1412772  ....  3.1417504 

512 3.1415138  ....  3.1416321 

1024 3.1415729  ....  3.1416025 

2048 3.1415877  ....  3.1415951 

4096 3.1415914  ....  3.1415933 

8192 3.1415923  ....  3.1415928 

16384 3.1415925  ....  3.1415927 

32768 3.1415926  ....  3.1415926 

The  area  of  the  circle,  we  infer  therefore,  is  equal  to 
3.1415926.  Some  doubt  may  exist  perhaps  about  the  last  de- 
cimal figure,  owing  to  errors  proceeding  from  the  parts  omitted ; 
but  the  calculation  has  been  carried  on  with  an  additional 
figure,  that  the  final  result  here  given  might  be  absolutely  cor- 
rect even  to  the  last  decimal  place. 

Since  the  area  of  the  circle  is  equal  to  half  the  circumfe- 
rence multiplied  by  the  radius,  the  half  circumference  must  be 
3.1415926,  when  the  radius  is  1  ;  or  the  whole  circumference 
must  be  3.1415926,  when  the  diameter  is  1  :  hence  the  ratio 
of  the  circumference  to  the  diameter,  formerly  expressed  by  tt, 
is  equal  to  3.1415926.  The  number  3.1416  is  the  one  gene- 
rally used.  I^* 


126  GEOMETRY. 


V 


BOOK  VI. 


PLANES  AND  SOUD  ANGLES. 


Definitions, 

1.  A  straight  line  is  perpendicular  to  a  plane,  when  it  is  per- 
pendicular to  all  the  straight  lines  which  pass  through  its  foot 
in  the  plane.  Conversely,  the  plane  is  perpendicular  to  the 
line. 

The  foot  of  the  perpendicular  is  the  point  in  which  the  per- 
pendicular line  meets  the  plane. 

2.  A  line  is  parallel  to  a  plane,  when  it  cannot  meet  thai 
plane,  to  whatever  distance  both  be  produced.  Conversely, 
the  plane  is  parallel  to  the  line. 

3.  Two  planes  are  parallel  to  each  other,  when  they  cannot 
meet,  to  whatever  distance  both  be  produced. 

4.  The  angle  or  mutual  inclination  of  two  planes  is  the  quan- 
tity, greater  or  less,  by  which  they  separate  from  each  other ; 
this  angle  is  measured  by  the  angle  contained  between  two 
lines,  one  in  each  plane,  and  both  perpendicular  to  the  common 
intersection  at  the  same  point. 

This  angle  may  be  acute,  obtuse,  or  a  right  angle. 
If  it  is  a  right  angle,  the  two  planes  are  perpendicular  to 
each  other. 

5.  A  solid  angle  is  the  angular  space  in-  S 
eluded  between  several  planes  which  meet  ,  y/f^ 
at  the  same  point.                                                        yy    l  \ 

Thus,  the  solid  angle  S,  is  formed  by  ..^  /Y  /  1 
the  union  of  the  planes  ASB,  BSC,  CSD,         JJ^—f-^C 

^®^-                                              ..              /         / 
Three  planes  at  least,  are  requisite  to     ^ ^ 

form  a  solid  angle.  -^  ^ 


BOOK  VI.  127 

PROPOSITION  I.    THEOREM. 

A  straight  line  cannot  be  partly  in  a  plane,  and  partly  out  of  it. 

For,  by  the  definition  of  a  plane,  when  a  straight  li.ne  has 
two  points  common  with  a  plane,  it  lies  wholly  in  that  plane. 

Scholium.  To  discover  whether  a  surface  is  plane,  it  is  ne- 
cessary to  apply  a  straight  line  in  different  ways  to  that  sur- 
face, and  ascertain  if  it  touches  the  surface  throughout  its  whole 
extent. 

PROPOSITION  II.    THEOREM. 

Two  straight  lines,  which  intersect  each  other,  lie  in  the  same 
plane,  and  determine  its  position. 

Let  AB,  AC,  be  two  straight  lines  which 
intersect  each  other  in  A  ;  a  plane  may  be 
conceived  in  which  the  straight  Hne  AB  is 
found  ;  if  this  plane  be  turned  round  AB,  until 
it  pass  through  the  point  C,  then  the  line  AC, 
which  has  two  of  its  points  A  and  C,  in  this 
plane,  lies  wholly  in  it ;  hence  the  position  of 
the  plane  is  determined  by  the  single  condition  of  containing 
the  two  straight  lines  AB,  AC. 

Cor.  1.  A  triangle  ABC,  or  three  points  A,  B,  C,  not  in  a 
straight  line,  determine  the  position  of  a  plane. 

Cor.  2.  Hence  also  two  parallels 
AB,  CD,  determine  the  position  of  a 
plane ;  for,  drawing  the  secant  EF, 
the  plane  of  the  two  straight  lines 
AE,  EF,  is  that  of  the  parallels 
AB,  CD. 


PROPOSITION  III.    THEOREM. 

If  two  planes  cut  each  other,  their  common  intersection  will  be  a 
straight  line. 


128 


GEOMETRY. 


Let  the  two  planes  AB,  CD,  cut 
each  other.  Draw  tlie  straight  line 
EF,  joining  any  two  points  E  and  F  in 
the  common  section  of  the  two  planes. 
This  line  will  lie  wholly  in  the  plane 
AB,  and  also  wholly  in  the  plane  CD 
(Book  J.  Def.  6.) :  therefore  it  will  be 
in  both  planes  at  once,  and  conse- 
quently is  their  common  intersection. 


-,P=*C1 


»^.. 


.--B 


D 


■iS  PROPOSITION.  IV.    THEOREM. 

If  a  straight  line  be  perpendicular  to  two  straight  lines  at  their 
point  of  intersection^  it  will  be  perpendicular  to  the  plane  oj 
those  lines. 


Let  MN  be  the  plane  of  the 
two  lines  BB,  CC,  and  let  AP 
be  perpendicular  to  them  at 
their  point  of  intersection  P ; 
then  will  AP  be  perpendicular 
to  every  line  of  the  plane  pass- 
ing through  P,  and  consequently 
to  the  plane  itself  (Def.  1.). 

Through  P,  draw  in  the  plane 
MN,  any  straight  line  as  PQ, 
and  through  any  point  of  this 
line,  a:s  Q,  drawBQC,  eo  that  BQ  shall  be  equal  to  QC  (Book 
IV.  Prob.  V.) ;  draw  AB,  AQ,  AC. 

The  base  BC  being  divided  into  two  equal  parts  at  the  point 
Q,  the  triangle  BPC  will  give  (Book  IV.  Prop.  XIV.), 
PCHPB2=2PQH2QC2. 

The  triangle  BAC  will  in  like  manner  give, 

AC2+AB2=2AQ-+2QCl     - 

Taking  the  first  equation  from  the  second,  and  observing 
that  the  triangles  APC,  APB,  which  are  both  right  angled  at 
P,  give 

AC2— PC2= AP2,  and  AB^— PB2= AP^ ; 
we  shall  have 

AP2+AP2=2AQ2— 2PQ^. 

Therefore,  by  taking  the  halves  of  both,  we  have 
AP2=:AQ^— PQ2,  or  AQ^^AF+PQ^ ; 
hence  the  triangle  APQ  is  right  angled  at  P ;  hence  AP  is  per- 
pendicular to  PQ. 


BOOK  VI. 


120 


Schohupi.  Thus  it  is  evident,  not  only  tiiat  a  straight  line 
may  be  perpendicular  to  all  the  straight  lines  which  pass 
through  iis  foot  in  a  plane,  but  that  it  always  must  be  so,  when 
ever  it  is  perpendicular  to  two  straight  lines  drawn  in  the 
plana  ;  which  proves  the  first  Definition  to  be  accurate. 

Cor.  1.  The  perpendicular  AP  is  shorter  than  any  oblique 
line  AQ ;  therefore  it  measures  the  true  distance  from  the  point 
A  to  the  plane  MN. 

Cor.  2.  At  a  given  point  P  on  a  plane,  it  is  impossible  to 
erect  more  than  one  perpendicular  to  that  plane  ;  for  if  there 
could  be  two  perpendiculars  at  the  same  point  P,  draw  through 
these  two  perpendiculars  a  plane,  whose  intersection  with  the 
plane  MN  is  PQ ;  then  these  two  perpendiculars  would  be  per- 
pendicular to  the  line  PQ,  at  the  same  point,  and  in  the  same 
plane,  which  is  impossible  (Book  I.  Prop.  XIV.  Sch.). 

It  is  also  impossible  to  let  fall  from  a  given  point  out  of  a 
plane  two  perpendiculars  to  that  plane ;  for  let  AP,  AQ,  be 
these  tv:o  perpendiculars,  then  the  triangle  APQ  would  have 
two  right  angles  APQ,  AQP,  which  is  impossible. 


PROPOSITION  V.    THEOREM. 


If  from  a  point  without  a  plane,  a  perpendicular  be  drawn  to  the 
plane,  and  oblique  lines  be  drawn  to  different  points, 

1st.  Any  two  oblique  lines  equally  distant  from  the  perpendicular 
will  be  equal. 

2d.  Of  any  two  oblique  lines  unequally  distant  from  the  perpen- 
dicular,  the  more  distant  will  be  the  longer. 

Let  AP  be  perpendicular  to 
the  plane  MN  ;  AB,  AC,  AD, 
oblique  hues  equally  distant 
from  the  perpendicular,  and 
AE  a  line  more  remote  :  then 
will  AB-AC=AD;  and  AE 
will  be  greater  than  AD. 

For,  the  angles  APB,  APC, 
APD,  being  right  angles,  if  we 
suppose  the  distances  PB,  PC, 
PI),  to  be  equal  to  each  other,  the  triangles  APB,  APC,  APD, 
will  have  in  each  an  equal  angle  contained  by  two  equal  sides ; 
herefore  they  will  be  equal ;  hence  the  hypothenuses,  or  the 
oblique  lines  AB,  AC,  AD,  will  be  equ^l  to  each  other.  In  like 


130 


GEOMETRY. 


manner,  if  the  distance  PE  is  greater  than  PD  or  its  pqual  PB, 
the  obhque  hne  AE  will  evidently  be  greater  than  Ali,  or  its 
equal  AD. 

Cor.  All  the  equal  oblique 
lines,  AB,  AC,  AD,  &c.  termi- 
nate in  the  circumference  BCD, 
described  from  Pthe  foot  of  the 
perpendicular  as  a  centre ; 
therefore  a  point  A  being  given 
out  of  a  plane,  the  point  P  at 
which  the  perpendicular  let  fall 
from  A  would  meet  that  plane, 
may  be  found  by  marking  upon 
that  plane  three  points  B,  C,  D,  equally  distant  from  the  pomt  A, 
and  then  finding  the  centre  of  the  circle  which  passes  through 
these  points  ;  this  centre  will  be  P,  the  point  sought. 

Scholium.  The  angle  ABP  is  called  the  inclination  of  the 
oblique  line  AB  to  the  plane  MN  ;  which  inclination  is  evidently 
equal  with  respect  to  all  such  lines  AB,  AC,  AD,  as  are  eqtially 
distant  from  the  perpendicular  ;  for  all  the  triangles  ABP,  ACP, 
ADP,  &c.  are  equal  to  each  other. 


PROPOSITION  VI.    THEOREM. 


If  from  a  point  without  a  plane,  a  perpendicular  he  let  fall  on  the 
plane,  and  from  the  foot  of  the  perpendicular  a  perpendicular 
be  drawn  to  any  line  of  the  plane,  and  from  the  point  of  inter- 
section a  line  be  drawn  to  the  first  point,  this  latter  line  will  he 
perpendicular  to  the  line  of  the  plane. 

Let  AP  be  perpendicular  to  the 
plane  NM,  and  PD  perpendicular  to 
BC ;  then  will  AD  be  also  perpen- 
dicular to  BC. 

Take  DB=DC.  and  draw  PB,  PC, 
AB,  AC.  Since  DB-DC,  the  ob- 
lique  line  PB  — PC:  and  with  regard 
to  the  perpendicular  AP,  since  PB=: 
PC,  the  oblique  line  AB=:AC  (Prop. 
V.  Cor.)  ;  therefore  the  line  AD  has 
two  of  its  points  A  and  D  equally  distant  from  the  extremities 
B  and  C  ;  therefore  AD  is  a  perpendicular  to  BC,  at  its  middle 
point  D  (Book  I.  Prop.  XVI.  Cor.). 


Ei 


BOOK  VI. 


131 


Cor.  It  is  evident  likewise,  that  BC  is  perpendicular  to  the 
plane  APD,  since  BC  is  at  once  perpendicuiar  to  the  two 
straight  lines  AD,  PD. 

Scholium.  The  two  lines  AE,  BC,  afford  an  instance  of  two 
lines  which  do  not  meet,  because  they  are  not  situated  in  the 
same  plane.  The  shortest  distance  between  these  lines  is  the 
straight  line  PD,  which  is  at  once  perpendicular  to  the  line  AP 
and  to  the  line  BC.  The  distance  PD  is  the  shortest  distance 
between  them,  because  if  we  join  any  other  two  points,  such 
as  A  and  B,  we  shall  have  AB>AD,  AD>PD;  therefore 
AB>PD.  ^ 

The  two  lines  AE,  CB,  though  not  situated  in  the  same  plane, 
are  conceived  as  forming  a  right  angle  with  each  other,  because 
AE  and  the  line  drawn  through  one  of  its  points  parallel  to 
BC  would  make  with  each  other  a  right  angle.  In  the  same 
manner,  the  line  AB  and  the  line  PD,  which  represent  any  two 
straight  lines  not  situated  in  the  same  plane,  are  supposed  to 
form  with  each  other  the  same  angle,  which  would  be  formed 
by  AB  and  a  straight  line  parallel  to  PD  drawn  through  one 
of  the  points  of  AB. 


/ 


PROPOSITION  VII.    THEOREM. 


If  one  of  two  parallel  lines  he  perpendicular  to  a  plane,  the  othe? 
will  also  be  perpendicular  to  the  same  plane. 

Let  the  lines  ED,  AP,  be 
parallel ;  if  AP  is  perpen- 
dicular to  the  plane  NM, 
then  will  ED  be  also  per- 
pendicular to  it. 

Through  the  parallels  AP, 
DE,  pass  a  plane  ;  its  inter- 
section with  the  plane  MN 
will  be  PD ;  in  the  plane  MN 
draw  BC  perpendicular  to  PD,  and  draw  AD. 

By  the  Corollary  of  the  preceding  Theorem,  BC  is  perpen- 
dicular to  the  plane  APDE  ;  therefore  the  angle  BDE  is  a  right 
angle ;  but  the  angle  EDP  is  also  a  right  angle,  since  AP'is 
perpendicular  to  PD,  and  DE  parallel  to  AP  (Book  I.  Prop. 
XX.  Cor.  1.)  ;  therefore  the  line  DE  is  perpendicular  to  the 
two  straight  lines  DP,  DB  ;  consequently  it  is  perpendicular  to 
their  plane  MN  (Prop.  IV.). 


132 


GEOMETRY. 


Cor.  1.  Conversely,  if  the 
straight  Hnes  AP,  DE,  are 
perpendicular  to  the  same 
plane  MN,  they  will  be  par- 
allel ;  for  if  they  be  not  so, 
draw  through  the  point  D.  a 
,  line  parallel  to  AP,  this  par- 
allel will  be  perpendicular 
to  the  plane  MN  ;  therefore 
through  the  same  point  D  more  than  one  perpendicular  might 
be  erected  to  the  same  plane,  which  is  impossible  (Prop.  IV. 
Cor.  2.). 

Cor,  2.  Two  lines  A  and  B,  parallel  to  a  third  C,  are  par- 
allel to  each  other  ;  for,  conceive  a  plane  perpendicular  to  the 
line  C  ;  the  lines  A  and  B,  being  parallel  to  C,  will  be  perpen- 
dicular to  the  same  plane  ;  therefore,  by  the  preceding  Corol- 
lary, they  will  be  parallel  to  each  other. 

The  three  lines  are  supposed  not  to  be  in  the  same  plane  ; 
otherwise  the  proposition  would  be  already  known  (Book  I. 
Prop.  XXII.). 


PROPOSITION  VIII.    THEOREM. 


If  a  straight  line  is  parallel  to  a  straight  line  drawn  in  a  plane^ 
it  will  bp  parallel  to  that  plane. 


Let  AB  be  parallel  to  CD 
of  the  plane  NM  ;  then  will 
it  be'  parallel  to  the  plane 
NM. 

For,  if  the  line  AB,  which 
lies  in  the  plane  ABDC, 
could  meet  the  plane  MN, 
this  could  only  be  in  some 


M 


K 


point  of  the  line  CD,  the  .common  intersection  of  the  two 
planes :  but  AB  cannot  meet  CD,  since  they  are  parallel ; 
hence  it  will  not  meet  the  plane  MN ;  hence  it  is  parallel  to 
that  plane  (Def.  2.). 


PROPOSITION  IX.    THEOREM. 


1 


TiJDO  planes  which  are  perpendicular  to  the  same  straight  line, 
are  parallel  to  each  other. 


BOOK  VI. 


133 


• — r~ 


t2 


^ 


■x: 


K" 


Let  the  planes  NM,  QP,  be  per-  *•• 
pendiciilar  to  the  hne  AB,  then  will  P 
they  be  parallel. 

For,  if  they  can  meet  any  where, 
let  O  be  one  of  their  common 
points,  and  draw  OA,  OB  ;  the  line 
AB  which  is  perpendicular  to  the 

plane  MN,  is  perpendicular  to  the  \ IQ. 

straight  line  OA  drawn  through  its  foot  in  that  plane  ;  for  the 
same  reason  AB  is  perpendicular  to  BO  ;  therefore  OA  and  OB 
are  two  perpendiculars  let  fall  from  the  same  point  O,  upon 
the  same  straight  line ;  which  is  impossible  (Book  I.  Prop.  XIV.); 
therefore  the  planes  MN,  PQ,  cannot  meet  each  other ;  consf^? 
quently  they  are  parallel. 


•         PROPOSITION  X.    THEOREM. 

If  a  plane  cut  two  parallel  planes,  the  lines  of  intersection  will  he 

parallel. 


Let  the  parallel  planes  NM, 
QP,  be  intersected  by  the  plane 
EH  ;  then  will  the  lines  of  inter- 
section EF,  GH,  be  parallel. 

For,  if  the  lines  EF,  GH,  lying 
in  the  same  plane,  were  not  par- 
allel, they  would  meet  each  other 
when  produced ;  therefore,  the 
planes  MN,  PQ,  in  which  those 
lines  lie,  would  also  meet ;  and 
hence  the  planes  would  not  be 
parallel. 


M  E 


PROPOSITION  XI.    THEOREM. 


If  two  planes  are  parallel,  a  straight  line  which  is  perpendicular 
to  cne,  is  ulso  perpendicular  to  the  other. 


M 


134 


GEOMETRY. 


M 

i      Sr 


B 


Let  MN,  PQ,  be  two  parallel 
planes,  and  let  AB  be  perpendicu- 
lar to  NM  ;  then  will  it  also  be  per- 
pendicular to  QP.  -J 

Having  drawn  any  line  BC  in 
the  plane  PQ,  through  the  lines  AB 
and  BC,  draw  a  plane  ABC.  inter- 
secting the  plane  MN  in  AD  ;  the 

intersection  AD  will  be  parallel  to  BC  (Prop.  X.)  ;  but  the  line 
AB,  being  perpendicular  to  the  plane  MN,  is  perpendicular  to 
the  straight  line  AD ;  therefore  also,  to  its  parallel  BC  (Book 
I.  Prop.  XX.  Cor.  1.):  hence  the  line  AB  being  perpendicular 
to  any  line  BC,  drawn  through  its  foot  in  the  plane  PQ,  is  con- 
sequently perpendicular  to  that  plane  (Def.  1.).  j 


ia 


PROPOSITION  XII.    THEOREM. 


The  parallels  comprehended  between  two  parallel  planes  are 

equal. 


Let  MN,  PQ,  be  two  parallel 
planes,  and  FH,  GE.  two  paral- 
lel lines  ;  then  will  EG=FH 

For,  through  the  parallels  EG, 
FH,  draw  the  plane  EGHF,  in- 
tersecting the  parallel  planes  in 
EF  and  GH.    The  intersections 

EF,  GH,  are  parallel  to  each 
other  (Prop.  X.) ;  so  likewise  are 

EG,  FH  ;  therefore  the  figure 
.kEGHF  is  a  parallelogram ;  con- 
sequently, EG  =FH. 


Cor.  Hence  it  follows,  that  two  parallel  planes  are  every 
where  equidistant :  for,  suppose  EG  were  perpendicular  to  the 
plane  PQ  ;  the  parallel  FH  would  also  be  perpendicular  to  it 
(Prop.  VII.),  and  the  two  parallels  would  likewise  be  perpen- 
dicular to  the  plane  MN  (Prop.  XL)  ;  and  being  parallel,  tliey  js 
will  be  equal,  as  shown  by  the  Proposition. 


BOOK  VI. 


135 


PROPOSITION  XIII.    THEOREM. 

If  two  angles,  not  situated  in  the  same  plane,  have  their  sides 
parallel  and  lying  in  the  same  direction,  those  angles  ivill  le 
eqhal  and  their  planes  will  he  parallel 

Let  the  angles  be  CAE  and  DBF. 

Make  AC-BD,  AE=  JM 
BF  ;  and  draw  CE,  DF, 
AB,  CD,  EF.  Since  AC 
is  equal  and  parallel  to 
BD,  the  figure  ABDC  is 
a  parallelogram ;  therefore 
CD  is  equal  and  parallel 
to  AB.  For  a  similar  rea- 
son, EF  is  equal  and  par- 
allel to  AB ;  hence  also  CD 
is  equal  and  parallel  to 
EF  ;  hence  the  figure 
CEFD  is  a  parallelogram, 
and  the  side  CE  is  equal 
and  paiallel  to  DF;  therefore  the  triangles  CAE, DBF,  have 
their  corresponding  sides  equal;  therefore  the  angle  CAE  — 
DBF. 

Again,  the  plane  ACE  is  parallel  to  the  plane  BDF.  For 
suppose  the  plane  drawn  through  the  point  A,  parallel  to  BDF, 
were  to  meet  the  lines  CD,  EF,  in  points  different  from  C  and 
E,  for  instance  in  G  and  H  ;  then,  the  three  lines  AB,  GD,  FH, 
would  be  equal  (Prop.  XII.) :  but  the  lines  AB,  CD,  EF,  are 
already  known  to  be  equal;  hence  CD=GD,  and  FH=EF, 
which  is  absurd  ;  hence  the  plane  ACE  is  parallel  to  BDF. 

Cor.  If  two  parallel  planes  MN,  PQ  are  met  by  two  other 
planes  CABD,  EABF,  the  angles  CAE,  DBF,  formed  by  the 
mtersections  of  the  parallel  planes  will  be  equal ;  for,  the  inter- 
section AC  is  parallel  to  BD,  and  AE  to  BF  (Prop.  X.) ;  there- 
fore the  angle  CAE = DBF. 


PROPOSITION  XIV.    THEOREM. 


If  three  straight  lines,  not  situated  in  the  same  plane,  are  equal 
and  parallel,  the  opposite  triangles  formed  by  joining  the  ex- 
tremities of  these  lines  will  be  equal,  and  their  planes  will  he 


i 


parillel 


136 


GEOMETRY. 


Let  AB,  CD,  EF,  be  the 
lines. 

Since  AB  is  equal  and 
parallel  to  CD,  the  figure 
ABDC  is  a  parallelogram  ; 
hence  the  side  AC  is  equal 
and  parallel  to  BD.  For  a 
like  reason  the  sides  AE, 
BF,  are  equal  and  parallel, 
as  also  CE,  DF ;  therefore 
the  two  triangles  ACE,  BDF, 
are  equa^ ;  hence,  by  the  last 
Proposition,  their  planes  are 
parallel. 


-M. 


C 

H 

^ 

/\G^ 

\.\-r. 

\ 

A^ 

-^.E 

A 

\    \ 

\ 

N 

P 

\    \ 

\?^ 

^\ 

\ 

B 

1? 

PROPOSITION  XV.    THEOREM. 


If  two  straight  lines  he  cut  hy  three  parallel  planes,  they  will  he 
divided  proportionally. 


Suppose  the  line  AB  to  meet 
the  parallel  planes  MN,  PQ, 
RS,  at  the  points  A,  E,  B ;  and 
the  line  CD  to  meet  the  same 
planes  at  the  points  C,  F,  D  : 
we  are  now  to  show  that 
AE  :  EB  :  :  CF  :  FD. 
Draw  AD  meeting  the  plane 
PQ  in  G,  and  draw  AC,  EG, 
GF,  BD ;  the  intersections  EG, 
BD,  of  the  parallel  planes  PQ, 
RS,  by  the  plane  ABD,  are 
parallel  (Prop.  X.)  ;  therefore 

AE  :  EB  :  :  AG  :  GD  ; 
in  like  manner,  the  intersections  AC,  GF,  being  parallel, 

AG  :  GD  :  :  CF  :  FD  ; 
the  ratio  AG  :  GD  is  the  same  in  both ;  hence 

AE  :  EB  :  :  CF  :  FD. 


1^ 

Y^ 

I 

M 

I^ 

^4^ 

.R 

^\v^ 

b' — — -^_^     • 

PROPOSITION  XVI.    THEOREM. 


Ifalt^  )  is  perpendicular  to  a  plane,  every  plane  passed  through 
thf     zrpendicular,  will  also  he.peipendicular  tojhe  plane. 


BOOK  VI. 


137 


Let  AP  be  perpen^ular  to  the 
plane  NM  ;  then  will  every  plane 
passing  through  AP  be  perpendicu- 
lar to  NM. 

Let  BC  be  the  intersection  of  the 
planes  AB,  MN  ;  in  the  plane  MN, 
draw  DE  perpendicular  to  BP :  then 
the  line  AP,  being  perpendicular  to 
the  plane  MN,  will  be  perpendicu- 
lar to  each  of  the  two  straight  lines 
BC,  DE  ;  but  the  angle  APD,  formed  by  the  two  perpendicu- 
lars PA,  PD,  to  the  common  intersection  BP,  measures  the 
angle  of  the  two  planes  AB,  MN  (Def.  4.) ;  therefore,  since  that 
angle  is  a  right  angle,  the  two  planes  are  perpendicular  to  each 
other. 

Scholium,  When  three  straight  lines,  such  as  AP,  BP,  DP, 
are  perpendicular  to  each  other,  each  of  those  lines  is  perpen- 
dicular to  the  plane  of  the  other  two,  and  the  three  planes  are 
perpendicular  to  each  other. 


PROPOSITION  XVII.    THEOREM. 


If  two  planes  are  perpendicular  to  each  other,  a  line  drawn  in 
one  of  them  perpendicular  to  their  common  intersectionj  .will 
be  perpendicular  to  the  other  plane,     ' 

Let  the  plane  AB  be  perpen- 
dicular to  NM  ;  then  if  the  line 
AP  be  perpendicular  to  the  inter- 
section BC,  it  will  also  be  perpen- 
dicular to  the  plane  NM. 

For,  in  the  plane  MN  draw  PD 
perpendicular  to  PB ;  then,  be- 
cause the  planes  are  perpendicu- 
lar, the  angle  APD  is  a  right  an- 
gle ;  therefore,  the  line  AP  is 
perpendicular  to  the  two  straight 

lines  PB,  PD ;  therefore  it  is  perpendicular  to  their  plane  MN 
(Prop.  IV.). 

Cor,  If  the  plane  AB  is  perpendicular  to  the  plane  MN,  and 
if  at  a  point  P  of  the  common  intersection  we  erect  a  perpen- 
dicular to  the  plane  MN,  that  perpendicular  will  be  in  the  plane 
AB  ;  for,  if  not,  then,  in  the  plane  AB  we  might  draw  AP  per- 

M* 


138 


GEOMETRY. 


pendicular  to  PB  the  common  intersection,  and  this  AP,  at  the 
same  time,  would  be  perpendicular  to  the  plane  MN;  therefore 
at  the  same  point  P  there  would  be  two  perpendiculars  to  the 
plane  MN,  which  is  impossible  (Prop.  IV.  Cor.  2.). 


PROPOSITION  XVIII.    THEOREM. 

Ij  two  planes  are  perpendicular  to  a  third  plane,  their  common 
intersection  will  also  he  peipendicular  to  the  third  plane. 


Let  the  planes  AB,  AD,  be  per- 
pendicular to  roi;  then  will  their 
intersection  AP  be  perpendicular 
toNM. 

For,  at  the  point  P,  erect  a  per- 
pendicular to  the  plane  MN" ;  that 
perpendicular  must  be  at  once  in 
the  plane  AB  and  in  the  plane  AD 
(Prop.  XVII.  Cor.)  ;  therefore  it 
is  theii  common  intersection  AP. 


PROPOSITION  XIX.    THEOREM. 


If  a  solid  angle  isfonmd  by  three  plane  angles,  the  sum  of  any 
two  of  these  angles  will  he  greater  than  the  third. 

The  proposition  requires  demonstra- 
tion only  when  the  plane  angle,  which 
is  compared  to  the  sum  of  the  other 
two,  is  greater  than  either  of  them. 
Therefore  suppose  the  solid  angle  S  to 
be  formed  by  three  plane  angles  ASB, 
ASC,  BSC,  whereof  the  angle  ASB  is 
the  greatest;  Ave  are  to  show  that 
ASB<ASC  +  BSC. 

In  the  plane  ASB  make  the  angle 
straight  line  ADB  at  pleasure;  and  having  taken  SC  =  SD, 
draw  AC,  BC. 

The  two  sides  BS,  SD,  are  equal  to  the  two  BS,  SC  ;  the 
angle  BSD=:BSC  ;  therefore  the  triangles  BSD,  BSC,  are 
equal;  therefore  BD=BC.  But  AB<AC  +  BC;  taking  BD 
from  the  one  side,  and  from  the  other  its  equal  BC,  there  re- 


BSD=:BSC,  draw  the 


139 

mains  AD<AC.  The  two  sides  AS,  SD,  are  equal  to  the 
two  AS,  SC  ;  the  third  side  AD  is  less  than  the  third  si-^e  AC  ; 
therefore  the  angle  ASD<ASC  (Book  I.  Prop.  IX.  Sch.). 
Adding  BSD=:=BSC,  we  shall  have  ASD  +  BSD  or  ASB< 
ASC  +  BSC. 


PROPOSITION  XX.    THEOREM. 

The  sum  of  the  plane  angles  which  form  a  solid  angle  is  always 
less  than  four  right  angles. 

Cut  the  solid  angle  S  by  any  plane  S 

ABCDE ;  from  O,  a  point  in  that  plane,  /A 

draw  to  the  several  angles  the  straight  //   \\ 

lines  AO,  OB,  OC,OD,bE.  //     \\ 

The  sum  of  the  angles  of  the  triangles  /  •'  /     l\ 

ASB,  BSC,  &c.  formed  about  the^vertex  /  ,.(M. ilry 

S,  is  equal  to  the  sum  of  the  angles  of  an        />-'''     */       .a\ 

equal  number  of  triangles  AOB,  BOC,  &c.   A.^- y- ■-)<o  1 1 

formed  about  the  point  O.     But  at  the  >.  //     xll 

point  B  the  sum  of  the  angles  ABO,  OBC,  ^^ ^ 

equal  to  ABC,  is  less  than  the  sum  of  the 
angles  ABS,  SBC  (Prop.  XIX.) ;  in  the  same  manner  at  the 
point  C  we  have  BCO  +  OCD<BCS  +  SCD;  and  so  with  all 
the  angles  of  the  polygon  ABCDE :  whence  it  follows,  that  the 
sum  of  all  the  angles  at  the  bases  of  the  triangles  whose  vertex 
is  in  O,  is  less  than  the  sum  of  the  angles  at  the  bases  of  the 
triangles  whose  vertex  is  in  S ;  hence  to  make  up  the  defi- 
ciency, the  sum  of  the  angles  formed  about  the  point  O,  is 
greater  than  the  sum  of  the  angles  formed  about  the  point  S. 
But  the  sum  of  the  angles  about  the  point  O  is  equal  to  four 
right  angles  (Book  I.  Prop.  IV.  Sch.) ;  therefore  the  sum  of  the 
plane  angles,  which  form  the  solid  angle  S,  is  less  than  four 
right  angles. 

Scholium.  This  demonstration  is  founded  on  the  supposition 
that  the  solid  angle  is  convex,  or  that  the  plane  of  no  one  sur- 
face produced  can  evej:  meet  the  solid  angle  ;  if  it  were  other- 
wise, the  sum  of  the  plane  angles  would  no  longer  be  limited, 
and  might  be  of  any  magnitude. 


PROPOSITION  XXI.    THEOREM. 

If  two  solid  angles  are  contained  by  three  plane  angles  which  are 
equal  to  each  other,  each  to  each,  the  planes  of  the  equal  angles 
will  be  equally  inclined  to  eachMher, 


140  GEOMETRY. 

Let  the  angle  ASC=DTF,the 
angle  ASB=DTE,  and  the  an- 
gle BSC=ETF;  then  will  the 
mclination  of  the  planes  ASC, 
ASB,  be  equal  to  that  of  the 
planes  DTF,  DTE. 

Having  taken  SB  at  pleasure, 
draw  BO  perpendicular  to  the 
plane  ASC ;  from  the  point  O,  at  which  the  perpendicular 
meets  the  plane,  draw  OA,  OC  perpendicular  to  SA,  SC  ; 
draw  AB,  BC  ;  next  take  TE  =  SB  ;  draw  EP  perpendicular  to 
the  plane  DTF ;  from  the  point  P  draw  PD,  PF,  perpendicular 
respectively  to  TD,  TF ;  lastly,  draw  DE,  EF. 

The  triangle  SAB  is  right  angled  at  A,  and  the  triangle  TDE 
at  D  (Prop.  VI.) ;  and  since  the  angle  ASB  =  DTE  we  have 
SBA=TED.  Likewise  SB = TE ;  therefore  the  triangle  SAB 
is  equal  to  the  triangle  TDE;  therefore  SA  =  TD,  and  AB  =  DE. 
In  like  manner,  it  may  be  shown,  that  SC=TF,  and  BC=EF. 
That  granted,  the  quadrilateral  SAOC  is  equal  to  the  quadri-    • 
lateral  TDPF:  for,  place  the  angle  ASC  upon  its  equal  DTF;    S 
because  SA=:TD,  and  SC=TF,  the  point  A  will  fall  on  D,.:  i 
and  the  point  C  on  F ;  and  at  the  same  time,  AO,  which  is  per- 
pendicular to  SA,  will  fall  on  PD  which  is  perpendicular  to 
TD,  and  in  like  manner  OC  on  PF ;  wherefore  the  point  O 
will  fall  on  the  point  P,  and  AO  will  be  equal  to  DP.    But  the 
triangles  AOB,  DPE,  are  right  angled  at  O  and  P ;  the  hypo- 
thenuse  AB=I)E,  and  the  side  AO=DP:  hence  those  trian-     ^ 
gles  are  equal  (Book  I.  Prop.  XVII.)  ;   and  consequently,  the    | 
angle  OAB  =  PDE.     The  angle  OAB  is  the  inclination  of  the     ' 
two  planes  ASB.  ASC  ;  and  the  angle  PDE  is  that  of  the  two 
planes  DTE,  DTF ;  hence  those  two  inclinations  are  equal  to 
each  other. 

It  must,  how^ever,  be  observed,  that  the  angle  A  of  the  right 
angled  triangle  AOB  is  properly  the  inclination  of  the  two 
planes  ASB,  ASC,  only  when  the  perpendicular  BO  falls  on 
the  same  side  of  SA,  with  SC  ;  for  if  it  fell  on  the  other  side, 
the  angle  of  the  two  planes  would  be  obtuse,  and  the  obtuse 
angle  together  with  the  angle  A  of  the  triangle  OAB  would 
make  two  right  angles.  But  in  the  same  case,  tl^e  angle  of  the 
two  planes  TDE,  TDF,  would  also  be  obtuse,  and  the  obtuse 
angle  together  with  the  angle  D  of  the  triangle  DPE,  would 
make  two  right  angles ;  and  the  angle  A  being  thus  always 
equal  to  the  angle  at  D,  it  would  follow  in  the  same  manner  that 
the  inclination  of  the  two  planes  ASB,  ASC,  must  be  equal  to 
that  of  the  two  planes  TDE,  TDF. 

Scholium.  If  two  solid  angles  are  contained  by  three  plane 


BOOK  VI.  141 

angles,  respectively  equal  to  each  other,  and  if  at  the  same  time 
the  equal  or  homologous  angles  are  disposed  in  the  same  man- 
ner in  the  two  solid  angles,  these  angles  will  be  equal,  and  they 
will  coincide  when  applied  the  one  to  the  other.  We  have 
already  seen  that  the  quadrilateral  SAOC  may  be  placed  upon 
its  equal  TDPF ;  thus  placing  SA  upon  TD,  SC  falls  upon  TF, 
and  the  point  O  upon  the  point  P.  Btit  because  the  triangles 
AOB,  DPE,  are  equal,  OB,  perpendicular  to  the  plane  ASC, 
is  equal  to  PE,  perpendicular  to  the  plane  TDF  ;  besides,  those 
perdendiculars  lie  in  the  same  direction ;  therefore,  the  point 
B  will  fall  upon  the  point  E,  the  line  SB  upon  TE,  and  the  two 
solid  angles  will  wholly  coincide. 

This  coincidence,  however,  tai^es  place  only  when  we  suj>- 
pose  that  the  equal  plane  angles  are  arranged  in  the  same  man- 
ner in  the  two  solid  angles  ;  for  if  they  were  arranged  in  an  in- 
verse order,  or,  what  is  the  same,  if  the  perpendiculars  OB,  PE, 
instead  of  lying  in  the  same  direction  with  regard  to  the  planes 
ASC,  DTF,  lay  in  opposite  directions,  then  it  would  be  impos- 
sible to  make  these  solid  angles  coincide  with  one  another.  It 
would  not,  however,  on  this  account,  be  less  true,  as  our  Theo- 
rem states,  that  the  planes  containing  the  equal  angles  must 
still  be  equally  inclined  to  each  other;  so  that  the  two  solid  an- 
gles would  be  equal  in  all  their  constituent'  parts,  without, 
however,  admitting  of  superposition.  This  sort  of  equalit}% 
which  is  not  absolute,  or  such  as  admits  of  superposition,  de- 
serves to  be  distinguished  by  a  particular  name  :  we  shall  call 
it  equality  hy  symmetry. 

Thus  those  two  solid  angles,  which  are  formed  by  three 
plane  angles  respectively  equal  to  each  other,  but  disposed  in  an 
inverse  order,  will  be  called  angles  equal  hy  symmetry ,  or  simply 
symmetrical  angles. 

The  same  remark  is  applicable  to  solid  angles,  which  are 
formed  by  more  than  three  plane  angles  :  thus  a  solid' angle, 
formed  by  the  plane  angles  A,  B,  C,  D,  E,  and  another  solid 
angle,  formed  by  the  same  angles  in  an  inverse  order  A,  E,  D, 
C,  B,  may  be  such  that  the  planes  which  contain  the  equal  an- 
gles are  equally  inclined  to  each  other.  Those  two  solid  angles, 
are  likew^ise  equal,  without  being  capable  of  superposition,  and 
are  called  solid  angles  equal  by  symmetry ,  or  symmetrical  solid 
angles. 

Among  plane  figures,  equality  by  symmetry  does  not  pro- 
perly exist,  all  figures  which  might  take  this  name  being  abso- 
lutely equal,  or  equal  by  superposition  ;  the  reason  of  which  is, 
that  a  plane  figure  may  be  inverted,  and  the  upper  part  taken 
indiscriminately  for  the  under.  This  is  not  the  case  with  solids ; 
in  which  the  third  dimension  may  be  taken  in  two  different 
directions. 


142 


GEOMETRY. 


BOOK  VII. 


POLYEDRONS. 

Definitions, 

1.  The  name  solid  pnlyedronf  or  simple  jjolyedron,  is  given 
to  every  solid  terminated  by  planes  or  plane  faces;  which 
planes,  it  is  evident,  will  themselves  be  terminated  by  straight 
lines. 

2.  The  common  intersection  of  two  adjacent  faces  of  a 
polyedron  is  called  the  side,  or  edge  of  the  polyedron. 

3.  The  prism  is  a  solid  bounded  by  several  parallelograms, 
which  are  terminated  at  both  ends  by  equal  and  parallel 
polygons. 

IC  7c 

T 


To  construct  this  solid,  let  ABCDE  be  any  polygon  ;  then 
if  in  a  plane  parallel  to  ABCDE,  the  lines  FG,  GH,*HI,  &c.  be 
drawn  equal  and  parallel  to  the  sides  AB,  BC,  CD,  &c.  thus 
forming  the  polygon  FGHIK  equal  to  ABCDE  ;  if  in  the  next 
place,  the  vertices  of  the  angles  in  the  one  plane  be  joined  with 
the  homologous  vertices  in  the  other,  by  straight  lines,  AF,  BG, 
CPI,  &c.  the  faces  ABGR  BCHG,  &c.  will  be  parallelograms, 
and  ABCDE-K,  the  solid  so  formed,  will  be  a  prism. 

4.  The  equal  and  parallel  polygons  ABCDE, 'FGHIK,  are 
called  the  bases  of  the  prism;  the  parallelograms  taken  together 
constitute  the  lateral  or  convex  surface  of  the  prism;  the  equal 
straight  lines  AF,  BG,  CH,  &c.  are  called  the  sides,  or  edges  of 
the  prism. 

5.  The  altitude  of  a  prism  is  the  distance  between  its  two 
bases,  or  the  perpendicular  drawn  from  a  point  in  the  upper 
base  to  the  plane  of  the  lower  base. 


BOOK  VII. 


143 


6.  A  prism  is  right,  when  the  sides  AF,  BG,  CH,  &;c.  are 
perpendicular  to  the  planes  of  the  bases ;  and  then  each  of  them 
is  equal  to  the  altitude  of  the  prism.  In  every  other  case  the 
prism  is  oblique,  and  the  altitude  less  than  the  side. 

7.  A  prism  is  triangular,  quadrangular,  pentagonal,  hex- 
agonal,  &c.  when  the  base  is  a  triangle,  a  quadrilateral,  a 
pentagon,  a  hexagon,  &c. 

8.  A  prism  whose  base  is  a  parallelogram,  and 

which  has  all  its  faces  parallelograms,  is  named  a 
parallelopipedon. 

The  parallelopipedon  is  rectangular  when  all 
its  faces  are  rectangles. 

9.  Among  rectangular  parallelopipedons,  we 
distinguish  the  cube,  or  regular  hexaedron, bounded 
by  six  equal  squares. 

10.  A  pyramid  is  a  solid  formed  by 
several  triangular  planes  proceeding  from 
the  same  point  S,  and  terminating  in  the 
different  sides  of  the  same  polygon 
ABCDE. 

The  polygon  ABCDE  is  called  the 
base  of  the  pyramid,  the  point  S  the 
vertex ;  and  the  triangles  ASB,  BSC, 
CSD,  &c.  form  its  convex  or  lateral  sur- 

11.  If  from  the  pyramid  S-ABCDE, 
the  pyramid  S-abcde  be  cut  off  by  a 
plane  parallel  to  the  base,  the  remaining 
solid  ABCDE-c?,  is  called  a  truncated 
pyramid,  or  the  frustum  of  a  pyramid. 

12.  The  altitude  of  a  pyramid  is  the 
perpendicular  let  fall  from  the  vertex  upon 
base,  produced  if  necessary. 

IS.  A  pyramid  is  triangular,  quadrangular,  &c.  according 
as  its  base  is  a  triangle,  a  quadrilateral,  &c. 

14.  A  pyramid  is  regular,  when  its  base  is  a  regular  poly- 
gon, and  when,  at  the  same  time,  the  perpendicular  let  fall 
from  the  vertex  on  the  plane  of  the  base  passes  through  the 
centre  of  the  base.  That  perpendicular  is  then  called  the  axis 
of  the  pyramid. 

15.  Any  line,  as  SF,  drawn  from  the  vertex  S  of  a  regular 
pyramid,  perpendicular  to  either  side  of  the  polygon  w^hich 
forms  its  base,  is  called  the  slant  height  of  the  pyramid. 

16.  The  diagonal  of  a  polyedron  is  a  straight  line  joining 
the  vertices  of  two  solid  angles  which  are  not  adjacent  to  each 
other. 


A 

the  plane  of  the 


144    ^ 


GEOMETRY. 


17.  Two  polyedrons  are  similar  when  they  are  contained 
by  the  same  number  of  similar  planes,  similarly  situated,  and 
having  like  inclinations  with  each  other. 


PROPOSITION  I.     THEOREM. 


The  convex  surface  of  a  right  prism  is  equal  to  the  perimeter  oj 
its  base  multiplied  hy  its  altitude. 

liCt  ABCDE-K  be  a  right  prism :  then 
will  its  convex  surface  be  equal  to 
(AB  +  BC  +  CD  +  DE  +  EA)  x  AF. 

For,  the  convex  surface  is  equal  to  the 
sum  of  all  the  rectangles  AG,  BH,  CI, 
DK,  EF,  which  compose  it.  Now,  the 
altitudes  AF,  BG,  CH,  &c.  of  the  rect- 
angles, are  equal  to  the  altitude  of  the 
prism.  Hence,  the  sum  of  these  rectan- 
gles, or  the  convex  surface  of  the  prism, 
is  equal  to  (AB  +  BC  +  CD  +  DE  +  EA)  x 
AF ;  that  is,  to  the  perimeter  of  the  base  of  the  prism  multi 
plied  by  its  altitude. 

Cor.    If  two  right  prisms  have  the  same  altitude,  their  con- 
vex surfaces  will  be  to  each  other  as  the  perimeters  of  theii    ; 
bases 


PROPOSITION  II.    THEOREM. 


In  every  prism,  the  sections  formed  hy  parallel  planes,  are  equal 

polygons.  ^ 


Let  the  prism  AH  be  intersected  by 
the  parallel  planes  NP,  S'V  ;  then  are  the 
polygons  NOPQR,  STVXY  equal. 

For,  the  sides  ST,  NO,  are  parallel, 
being  the  intersections  of  two  parallel 
planes  with  a  third  'plane  ABGF ;  these 
same  sides,  ST,  NO,  are  included  between 
the  parallels  NS,  OT,  which  are  sides  of 
the  prism:  hence  NO  is  equal  to  ST. 
For  like  reasons,  the  sides  OP,  PQ,  QR, 
&c.  of  the  section  NOPQR,  are  equal 
to  the  sides  TV,  VX,  XY,  <&c.  of  the  sec- 
tion STVXY,  each  to  each.    And  since 


BOOK  VII. 


145 


the  equal  sides  are  at  the  same  time  parallel,  it  follows  that  the 
angles  NOP,  OPQ,  &c.  of  the  first  section,  are  equal  to  the 
angles  STV,TVX,  &c.  of  the  second,  each  to  each  (Book  VI. 
Prop.  XIIL).  Hence  the  two  sections  NOPQR,  STVXY,  are 
equal  polygons. 

Cor.  Every  section  in  a  prism,  if  drawn  parallel  to  the  base, 
is  also  equal  to  the  base. 


PROPOSITION  III.    THEOREM. 


If  a  pyramid  he  cut  by  a  plane  parallel  to  its  base, 

1st.  The  edges  and  the  altitude  will  be  divided  proportionally. 

2d.  The  section  will  be  a  polygon  similar  to  tlie  base. 

Let  the  pyramid  S-ABCDE, 
of  which  SO  is  the  altitude, 
be  cut  by  the  plane  abcde ; 
then  will  Sa  :  SA  :  :  So  :  SO, 
and  the  same  for  the  other 
edges :  and  the  polygon  abcde, 
will  be  similar  to  the  base 
ABCDE. 

First.  Since  the  planes  ABC, 
abc,  are  parallel,  their  intersec- 
tions AB,  ab,  by  a  third  plane 
SAB  will  also  be  parallel 
(Book  VI. 'Prop.  X.) ;  hence  the  triangles  SAB,  Sab  are  simi- 
lar, and  we  have  SA  :  Sa  :  :  SB  :  S6 ;  for  a  similar  reason, 
we  have  SB  :  S6  :  :  SC  :  Sc;  and  so  on.  Hence  the  edges 
SA,  SB,  SC,  &:c.  are  cut  proportionally  in  a,  6,  c,  &c.  The 
altitude  SO  is  likewise  cut  in  the  same  proportion,  at  the  point 
o ;  for  BO  and  bo  are  parallel,  therefore  we  have 
SO  :  So  :  :  SB  :  Sfe. 

Secondly.  Since  ab  is  parallel  to  AB,  be  to  BC,  cd  to  CD,  &c. 
the  angle  abc  is  equal  to  ABC,  the' angle  bed  to  BCD,  and  so  on 
(Book  VI.  Prop.  XHL).  Also,  by  reason  of  the  similar  trian- 
gles SAB,  S«6,  we  have  AB  :  ab  :  :  SB  :  S6  ;  and  by  reason 
of  the  similar  triangles  SBC,  Sbc,  we  have  SB  :  Sb  :  :  BC  : 
be ;  hence  AB  :  ab  :  :  BC  :  be ;  we  might  likewise  have 
BC  :bc  :  :  CD  :  cd,  and  so  on.  Hence  the  polygons  ABCDE, 
abcde  have  their  angles  respectively  equal  and  their  homolo- 
gous sides  proportional ;  hence  they  are  similar. 

N 


146 


GEOMETRY. 


Cor.  1.  Let  S-ABCDE, 
S-XYZ  be  two  pyramids,  hav- 
ing a  common  vertex  and  the 
same  altitude,  or  having  their 
bases  situated  in  the  same 
plane  ;  if  these  pyramids  are 
cut  by  a  plane  parallel  to  the 
plane  of  their  bases,  giving  the 
sections  abcde,  xyz,  then  will 
the  sections  abcde,  xyz,  he  to  each 
other  as  the  bases  ABCDE, 
XYZ. 

For,  the  polygons  ABCDE,  abcde,  being  similar,  their  sur- 
faces are  as  the  squares  of  the  homologous  sides  AB,  ab  ;  but 
AB  :  «&  :  :  SA  :  S«;  hence  ABCDE  :  abcde  :  :  SA^  :  ^a\ 
For  the  same  reason,  XYZ  :  xyz  :  :  SX^  :  Sx^.  But  since 
abc  and  xyz  are  in  one  plane,  we  have  likewise  SA  :  Saj  :  : 
SX  :  So;  (Book  VI.  Prop.  XV.) ;  hence  ABCDE  :  abcde  :  : 
XYZ  :  xyz  ;  hence  the  sections  abcde,  xyz,  are  to  each  othei 
as  the  bases  ABCDE,  XYZ. 

Cor.  2.  If  the  bases  ABCDE,  XYZ,  are  equivalent,  any  sec-' 
tions  abcde,  xyz,  made  at  equal  distances  from  the  bases,  will 
be  equivalent  likewise. 


PROPOSITION  IV.    THEOREM. 


The  convex  surface  of  a  regular  "pyramid  is  equal  to  the  perime 
ter  of  its  base  multiplied  by  half  the  slant  height. 

For,  since  the  pyramid  is  regular,  the 
point  O,  in  which  the  axis  meets  the  base, 
is  the  centre  of  the  polygon  ABCDE 
(Def.  14.) ;  hence  thelines  OA,  OB,  OC, 
&c.  drawn  to  the  vertices  of  the  base, 
are  equal. 

In  the  right  angled  triangles  SAO,  SBO, 
the  bases  and  perpendiculars  are  equal : 
since  the  bypothenuses  are  equal :  and 
it  may  be  proved  in  the  same  way  that 
all  the  sides  of  the  right  pyramid  are 
equal.  The  triangles,  therefore,  which 
form  the  convex  surface  of  the  prism  are 
all  equal  to  each  other.  But  the  area  of 
either  of  these  triangles,  as  ES  A,  is  equal 


BOOK  VII. 


147 


to  its  base  EA  multiplied  by  half  the  perpendicular  SF,  which 
is  the  slant  height  of  the  pyramid  :  hence  the  area  of  all  the  tri- 
angles, or  the  convex  surface  of  the  pyramid,  is  equal  to  the 
perimeter  of  the  base  multiplied  by  half  the  slant  height. 

Cor.  The  convex  surface  of  the  frustum  of  a  regular  pyra- 
mid is  equal  to  half  the  perimeters  of  its  upper  and  lower  bases 
multiplied  by  its  slant  height. 

For,  since  the  section  abcde  is  similar  to  the  base  (Prop.  III.), 
and  since  the  base  ABCDE  is  a  regular  polygon  (Def.  14.),  it 
follows  that  the  sides  e«,  ab,  be,  cd  and  de  are  all  equal  to  each 
other.  Hence  the  convex  surface  of  the  frustum  ABCDE-c? 
is  formed  by  the  equal  trapezoids  EAae,  AB6a,  &c.  and  the 
perpendicular  distance  between  the  parallel  sides  of  either  of 
these  trapezoids  is  equal  to  Ff  the  slant  height  of  the  frustum. 
But  the  area  of  either  of  the  trapezoids,  as  AEea,  is  equal  to 
|(EA  +  ea)  xF/  (Book  IV.  Prop.  VII.) :  hence  the  area  of  all 
of  them,  or  the  convex  surface  of  the  frustum,  is  equal  to  half 
the  perimeters  of  the  upper  and  lower  bases  multiplied  by  the 
slant  height. 

PROPOSITION  V.     THEOREM. 

If  the  three  planes  which  form  a  solid  angle  of  a  prism,  are  equal 
to  the  three  planes  which  form  the  solid  angle  of  another  prism, 
each  to  each,  and  are  like  situated,  the  two  prisms  will  be  equal 
to  each  other. 


Let  the  base  ABCDE  be  equal  to  the  base  abcde,  the  paral- 
lelogram ABGF  equal  to  the  parallelogram  abgf  and  the  par- 
allelogram BCHG  equal  to  bchg-,  then  will  the  prism  ABCDE-K 
be  equal  to  the  prism  abcde-k. 


For,  lay  the  base  ABCDE  upon  its  equal  abcde  ;  these  two 
bases  will  coincide.     But  the  three  plane  angles  which  form 


148 


GEOMETRY 


the  solid  angle  B,  are  respectively  equal  to  the  three  plane 
angles,  which  form  the  solid  angle  b,  namely,  ABCrrc/ic, 
ABGrrra^^,  and  GBC  =gbc  ;  they  are  also  similarly  situated  . 
hence  the  solid  angles  B  and  bare  equal  (Book  YI.  Prop.  XXL 
Sch.) ;  and  therefore  the  side  BG  will  fall  on  its  equal  bg.  It 
is  likewise  evident,  that  by  reason  of  the  equal  parallelograms 
ABGF,  abgff  the  side  GF  will  fall  on  its  equal  gf,  and  in  the 
same  manner  GH  on  gh ;  hence,  the  plane  of  the  upper  base, 
FGHIK  will  coincide  with  the  plane  fghik  (Book  VI.  Prop.  II.).; 
K  7c 


But  the  two  upper  bases  being  equal  to  their  corresponding 
lower  bases,  are  equal  to  each  other :  hence  HI  will  coincide 
with  hi,  IK  with  ik,  and  KF  with  A/;  and  therefore  the  lateral 
faces  of  the  prisms  will  coincide :  therefore,  the  two  prisms 
coinciding  throughout  are  equal  (Ax.  13.). 

Co?\  Two  right  prisms,  which  have  equal  bases  and  equal  al- 
titudes, are  equal.  For,  since  the  side  AB  is  equal  to  ab,  and 
the  altitude  BG  to  bg,  the  rectangle  ABGF  will  be  equal  to 
abgf;  so  also  will  the  rectangle  BGHC  be  equal  to  bghc ;  and 
thus  the  three  planes,  which  form  the  solid  angle  B,  will  be 
equal  to  the  three,  which  form  the  solid  angle  b.  Hence  the 
two  prisms  are  equal. 


PROPOSITION  VI.    THEOREM. 


In  every  parallelopipedon   the  opposite  planes  are  equal  and 

parallel. 


By  the  definition  of  this  solid,  the  bases 
ABCD,  EFGH,  are  equal  parallelograms, 
and  their  sides  are  parallel :  it  remains 
only  to  show,  that  the  same  is  true  of  any 
two  opposite  lateral  faces,  such  as  AEHD, 
BFGC.  Now  AD  is  equal  and  parallel 
to  BC,  because  tiie  figure  ABCD  is  a  par- 


E 

H 

/  r 

.4/ 

V— "" 

v^ 

B 

c 

BOOK  VII.  149 

allelogram ;  for  a  like  reason,  AE  is  parallel  to  BF :  hence  the 
angle  DAE  is  equal  to  the  angle  CBF,  and  the  planes  DAE, 
CBF,  are  parallel  (Book  VI.  Prop.  XIII.)  ;  hence  also  the  par- 
allelogram DAEH  is  equal  to  the  parallelogram  CBFG.  In  the 
same  way.  it  might  be  shown  that  the  opposite  parallelograms 
ABFE,  DCGH,  are  equal  and  parallel. 

Cor.  1.  Since  the  parallelopipedon  is  a  solid  bounded  by  six 
planes,  whereof  those  lying  opposite  to  each  other  are  equal 
and  parallel,  it  follows  that  any  face  and  the  one  opposite  to  it, 
may  be  assumed  as  the  bases  of  the  parallelopipedon. 

Cor.  2.  The  diagonals  of  a  parallelopipedon  bisect  each 
other.  For,  suppose  two  diagonals  EC,  AG,  to  be  drawn  both 
through  opposite  vertices :  since  AE  is  equal  and  parallel  to 
CG,  the  figure  AEGC  is  a  parallelogram ;  hence  the  diagonals 
EC,  AG  will  mutually  bisect  each  other.  In  the  same  manner, 
we  could  show  that  the  diagonal  EC  and  another  DF  bisect 
each  other ;  hence  the  four  diagonals  will  mutually  bisect  each 
other,  in  a  point  which  may  be  regarded  as  the  centre  of  the 
parallelopipedon. 

Scholium.  If  three  straight  lines  AB,  AE,  AD,  passing 
through  the  same  point  A,  and  making  given  angles  with  each 
other,  are  known,  a  parallelopipedon  may  be  formed  on  those 
lines.  For  this  purpose,  a  plane  must  be  passed  through  the 
extremity  of  each  line,  and  parallel  to  the  plane  of  the  other 
two  ;  that  is,  through  the  point  B  a  plane  parallel  to  DAE, 
through  D  a  plane  parallel  to  BAE,  and  through  E  a  plane 
parallel  to  BAD.  The  mutual  intersections  of  these  planes  will 
form  the  parallelopipedon  required. 


PROPOSITION  VII.    THEOREM. 

The  two  triangular  prisms  into  which  a  parallelopipedon  is  di' 
vided  by  a  plane  passing  through  its  opposite  diagonal  edges, 
are  equivalent. 


N* 


150  GEOMETRY. 

Let  the  parallelopipedon  ABCD-H  be 
divided  by  the  plane  BDHFpassingthrough 
its  diagonal  edges  :  then  will  the  triangular 
prism  ABD-H  be  equivalent  to  the  trian- 
gular prism  BCD-H. 

Through  the  vertices  B  and  F,  draw  the 
planes  Bade,  Fehg,  at  right  angles  to  the 
side  BF,  the  former  meeting  AE,  DH,  CG, 
the  three  other  sides  of  the  parallelopipe- 
don, in  the  points  a,  d,  c,  the  latter  in  e,  A, 
g :  the  sections  Bade,  Fehg,  will  be  equal 
parallelograms.  They  are  equal,  because 
they  are  formed  by  planes  perpendicular  to  the  same  straight 
line,  and  consequently  parallel  (Prop.  II.) ;  they  are  parallelo- 
grams, because  aB,  dc,  two  opposite  sides  of  the  same  section, 
are  formed  by  the  meeting  of  one  plane  with  two  parallel 
planes  ABFE,  DCGH. 

For  a  like  reason,  the  figure  B^zeF  is  a  parallelogram ;  so  also 
are  BF^c,  cd/ig,  adhe,  the  other  lateral  faces  of  the  solid  Badc-g ; 
hence  that  solid  is  a  prism  (Def  6.) ;  and  that  prism  is  right, 
because  the  side  BF  is  perpendicular  to  its  base. 

But  the  right  prism  Badc-g  is  divided  by  the  plane  BH  into 
two  equal  right  prisms  Bad-h,  Bcd-h  ;  for,  the  base^JB^j^,  Bed, 
of  these  prisms  are  equal,  being  halves  of  the  same  parallel- 
ogram, and  they  have  the  common  altitude  BF,  hence  they  are 
equal  (Prop.  V.  Cor.). 

It  is  now  to  be  proved  that  the  oblique  triangular  prism 
ABD-H  will  be  equivalent  to  the  right  triangular  prism  Bad-h ; 
and  since  those  prisms  have  a  common  part  ABD-/i,  it  will 
only  be  necessary  to  prove  that  the  remaining  parts,  namely, 
the  solids  BaADi,  FeEH/«,  are  equivalent. 

Now,  by  reason  of  the  parallelograms  ABFE,  «BFe,  the 
sides  AE,  ae,  being  equal  to  their  .parallel  BF,  are  equal  to  each 
other;  and  taking  away  the  common  part  Ae,  there  remains 
AanrEe.     In  the  same  manner  we  could  prove  Dc?=HA. 

Next,  to  bring  about  the  superposition  of  the  two  solids 
BflADd/,  FfiEH/i,  let  us  place  the  base  Yeh  on  its  equal  Bad : 
the  point  e  falling  on  a,  and  the  point  h  on  d,  the  sides  eE,  AH, 
will  fall  on  their  equals  a  A,  dD,  because  they  are  perpendicu- 
lar to  the  same  plane  Bad.  Hence  the  two  solids  in  question 
will  coincide  exactly  with  each  other  ;  hence  the  oblique  prism 
BAD-H,  is  equivalent  to  the  rigf;t  one  Bad-h. 

In  the  same  manner  might  the  oblique  prism  BCD-H,  be 
proved  equivalent  to  the  right  prism  Bcd-h,  But  the  two  right 
prisms  Bad-h,  Bcd-h,  are  equal,  since  they  have  the  same  alti- 
tude BF,  and  since  their  bases  Bad,  Bde,  are  halves  of  the 
same  parallelogram  (Prop.  V.  Cor.).    Hence  the  two  trian- 


BOOK  VII.  151 

gular  prisms  BAD-H,  BDC-G,  being  equivalent  to  the  equal 
right  prisms,  are  equivalent  to  each  other. 

Cor,  Every  triangular  prism  ABD-HEF  is  half  of  the  paral- 
lelopipedon  AG  described  with  the  same  solid  angle  A,  and 
the  same  edges  AB,  AD,  AE. 


PROPOSITION  VIII.    THEOREM.  - 

If  two  parallelopipedons  have  a  common  hase,  and  their  upper 
bases  in  the  same  plane  and  between  the  same  parallels^  they 
will  be  equivalent. 

Let  the  parallelopipe- 
dons AG,  AL,  have  the 
common  base  AC,  and 
their  upper  bases  EG, 
MK,  in  the  same  plane, 
and  between  the  same 
parallels  HL,  EK ;  then 
will  they  be  equivalent. 

There  may  be  three 
cases,  according  as  EI  is 
greater,  less  than,  or  equal  to,  EF ;  but  the  demonstration  is 
the  same  for  all.  In  the  first  place,  then  we  shall  show  that 
the  triangular  prism  AEI-MDII,  is  equal  to  the  triangular 
prism  BFK-LCG. 

Since  AE  is  parallel  to  BF,  and  HE  to  GF,  the  angle  AEI 
=BFK,  HEI-GFK,  and  HEA=GFB.  Also,  since  EF  and 
IK  are  each  equal  to  AB,  they  are  equal  to  each  other.  To 
each  add  FI,  and  there  will  result  EI  equal  to  FK :  hence  the 
triangle  AEI  is  equal  to  the.triangle  BFK  (Bk.  I.  Prop.  V),  and 
the  paralellogram  EM  to  the  parallelogram  FL.  But  the  par- 
allelogram AH  is  equal  to  the  parallelogram  CF  (Prop.  VI) : 
hence,  the  three  planes  which  form  the  solid  angle  at  E  are 
respectively  equal  to  the  three  which  form  the  solid  angle  at 
F,  and  being  like  placed,  the  triangular  prism  AEI-M  is  equal 
to  the  triangular  prism  BFK-E. 

But  if  the  prism  AEI-M  is  taken  away  from  the  solid  AL, 
there  will  remain  the  parallelopipedon  BADC-L ;  and  if  the 
prism  BFK-L  is  taken  away  from  the  same  solid,  there  will 
remain  the  parallelopipedon  BADC-G ;  hence  those  two  paral- 
lelopipedons BADC-L,  BADC-G,  are  equivalent. 


152 


GEOMETRY. 


PROPOSITION  IX.    THEOREM. 


Two  parallelopipedons,  having  the  same  base  and  the  same  alti- 
tude, are  equivalent. 


Let  ABCD  be  the  com- 
mon base  of  the  two  par- 
allelopipedons  AG,  AL  ; 
since  they  have  the  same 
altitude,  their  upper  bases 
EFGH,IKLM,willbein 
the  same  plane.  Also  the 
sides  EF  and  AB  will  be 
equal  and  parallel,  as  well 
as  IK  and  AB  ;  hence  EF 
is  equal  and  parallel  to 
IK;  for  a  like  reason,  GF 
is  equal  and  parallel  to 
LK.     Let  the  sides  EF,  GH,  be  produced 


B 

and  likewise 


IM,  till  by  their  intersections  they  form  the  parallelogram 
NOPQ ;  this  parallelogram  will  evidently  be  equal  to  either 
of  the  bases  EFGH,  IKLM.  Now  if  a  third  parallelopipedon 
be  conceived,  having  for  its  lower  base  the  parallelogram 
ABCD,  and  NOPQ  for  its  upper,  the  third  parallelopipedon 
will  be  equivalent  to  the  parallelopipedon  AG,  since  with  the 
same  lower  base,  their  upper  bases  lie  in  the  same  plane 
and  between  the  same  parallels,  GQ,  FN  (Prop.  VIII.). 
For  the  same  reason,  this  third  parallelopipedon  will  also  be 
equivalent  to  the  parallelopipedon  AL ;  hence  the  two  paral- 
lelopipedons  AG,  AL,  which  have  the  same  base  and  the 
same  altitude,  are  equivalent. 


PROPOSITION  X.     THEOREM. 


Any  parallelopipedon  may  be  changed  into  an  equivalent  rectan 
gular  parallelopipedon  having  the  same  altitude   and 
equivalent  base. 


an 


BOOK  VII. 


153 


Let  AG  be  the  par- 
allelopipedon  proposed. 
From  the  points  A,  B,  C, 
D,drawAI,BK,CL,DM, 
perpendiculartothe  plane 
of  the  base  ;  you  will  thus 
form  the  parallelopipe- 
don  AL  equivalent  to 
AG,  and  having  its  late- 
ral faces  AK,  BL,  &c. 
rectangles.  Hence  if  the 
base  ABCD  is  a  rectan- 
gle, AL  will  be  a  rectan- 
gular parallelopipedon  equivalent  to  AG,  and  consequently, 
the  parallelopipedon  required.  But  if  ABCD  is  not  a  rectangle, 
draw  AO  and  BN  perpendicular  to  CD,  and  MQ  XP 

OQ  and  NP  perpendicular  to  the  base  ;  you 
will  then  have  the  solid  ABNO-IKPQ,  which 
will  be  a  rectangular  parallelopipedon :  for 
by  construction,  the  bases  ABNO,  and  IKPQ 
are  rectangles ;  so  also  are  the  lateral  faces, 
the  edges  AI,  OQ,  &c.  being  perpendicular 
to  the  plane  of  the  base  ;  hence  the  solid  AP 
is  a  rectangular  parallelopipedon.  But  the 
two  parallelopipedons  AP,  AL  may  be  con- 
ceived as  having  the  same  base  ABKl  and 
the  same  altitude  AO :  hence  the  parallelopipedon  AG,  which 
was  at  first  changed  into  an  equivalent  parallelopipedon  AL, 
is  again  changed  into  an  equivalent  rectangular  parallelopipe- 
don AP,  having  the  same  altitude  AI,  and  a  base  ABNO  equi- 
valent to  the  base  ABCD. 


PROPOSITION  XI.    THEOREM. 


Two  rectangular  parallelopipedons,  wJiich  have  the  same  base, 
are  to  each  other  as  their  altitudes. 


(F 


M 


K 


D 


\ 


154  GEOMETRY. 

Let  the  parallelopipedons  AG,  AL,  have  the  same  base  ED; 
then  will  they  be  to  each  other  as  their  altitudes  AE,  AT. 

•First,  suppose  the  altitudes  AE,  AI,  to  be    j;  .H 

to  each  other  as  two  whole  numbers,  as  15  is 
to  8,  for  example.  Divide  AE  into  15  equal 
parts ;  whereof  AI  will  contain  8 ;  and  through  q 
X,  y,  z,  &c.  the  points  of  division,  draw  planes  ^\in 
parallel  to  the  base.  These  planes  will  cut 
the  solid  AG  into  15  partial  parallelopipedons, 
all  equal  to  each  other,  because  they  have  55.. 
equal  bases  and  equal  altitudes — equal  bases,  ^'. 
since  every  section  MIKL,  made  parallel  to 
the  base  ABCD  of  a  prism,  is  equal  to  that 
base  (Prop.  II.),  equal  altitudes,  because  the 
altitudes  are  the  equal  divisions  Ax,  xy,  yz, 
&c.  But  of  those  15  equal  parallelopipedons,  8  are  con- 
tained in  AL  ;  hence  the  solid  AG  is  to  the  solid  AL  as  15  is  to 
8,  or  generally,  as  the  altitude  AE  is  to  the  altitude  AI. 

Again,  if  the  ratio  of  AE  to  AI  cannot  bfe  exactly  expressed 
in  numbers,  it  is  to  be  shown,  that  notwithstanding,  we  shall 
have 

solid  AG  :  solid  AL  :  :  AE  :  AI. 
For,  if  this  proportion  is  not  correct,  suppose  we  have 

sol.  AG  :  soL  AL  :  :  AE  :  AO  greater  than  AI. 
Divide  AE  into  equal  parts,  such  that  each  shall  be  less  than 
01 ;  there  will  be  at  least  one  point  of  division  m,  between  O 
and  I.  Let  P  be  the  parallelopipedon,  whose  base  is  ABCD, 
and  altitude  Am  ;  since  the  altitudes  AE,  Am,  are  to  each  other 
as  the  two  whole  numbers,  we  shall  have 

50/.  AG  :  P  :  ;  AE  :  Am. 
But  by  hypothesis,  w^e  have 

sol  AG  :  sol  AL  :  :  AE  :  AO ; 
therefore, 

sol  AL  :  P  :  :  AO  :  Am. 
But  AO  is  greater  than  Am ;  hence  if  the  proportion  is  correct, 
the  solid  AL  must  be  greater  than  P.     On  the  contrary,  how- 
ever, it  is  less :  hence  the  fourth  term  of  this  proportion 

sol  AG  :  sol  AL  :  :  AE  :  x, 
cannot  possibly  be  a  line  greater  than  AI.  By  the  same  mode 
of  reasoning,  it  might  be  shown  that  the  fourth  term  cannot  be 
less  than  AI  ;  therefore  it  is  equal  to  AI ;  hence  rectangular 
parallelopipedons  having  the  same  base  are  to  each  other  as 
their  altitudes. 


BOOK  VII. 


155 


PROPOSITION  XII.    THEOREM. 

Two  rectangular  parallelopipedons,  having  the  same  altitude 
are  to  each  other  as  their  bases. 


1 


A 


I 


^ 


Let  the  parallelopipedons      x  E  11 

AG,  AK,  have  the  same  al- 
titude AE ;  then  will  they  be        \K_ 
to  each  other  as  their  bases 
AC,  AN. 

Having  placed  the  tvv^o  "^ |-HG 

solids  by  the  side  of  each 
other,  as  ithe  figure  repre- 
sents, produce  the  plane 
ONKL  till  it  meets  the 
plane  DCGH  in  PQ ;  you 
will  thus  have  a  third  par- 
allelopipedon  AQ,  which 
may  be  compared  with  each 
of  the  parallelopipedons 
AG,  AK.  The  two  solids 
AG,  AQ,  having  the  same 
base  AEHD  are  to  each  other  as  their  altitudes  AB,  AG  ;  in 
like  manner,  the  two  solids  AQ,  AK,  having  the  same  base 
AOLE,  are  to  each  other  as  their  altitudes  AD,  AM.  Hence 
we  have  the  two  proportions, 

sol.  AG  :  sol  AQ  :  :  AB  :  AG, 
sol  AQ  :  sol  AK  :  :  AD  :  AM. 
Multiplying  together  the  corresponding  terms  of  these  propor- 
tions, and  omitting  in  the  result  the  common  multiplier  sol  AQ ; 
we  shall  have 

50/.  AG  :  sol  AK  :  :  ABxAD  :  AOxAM. 
But  AB  X  AD  represents  the  base  ABCD  ;  and  AO  x  AM  rep- 
resents the  base  AMNO  ;  hence  two  rectangular  parallelopipe- 
dons of  the  same  altitude  are  to  each  other  as  their  bases. 


PROPOSITION  XIII.    THEOREM. 


Any  two  rectangular  parallelopipedons  are  to  each  other  as  the 
products  of  their  bases  by  their  altitudes,  that  is  to  say,  as  the 
products  of  their  three  dimensions. 


156 


GEOMETRY. 


E 


•% 


K 


V 


X 


Z 


M 


H 


T? 


^>. 


D 


For,  having  placed  the  two 
soHds  AG,  AZ,  so  that  their 
surfaces  have  the  common 
angle  BAE,  produce  the 
planes  necessary  for  com- 
pleting tlie  third  parallelopi- 
pedon  AK  having  the  same 
altitude  vt^ith  the  parallelopi- 
pedon  AG.  By  the  last  propo- 
sition, v^^e  shall  have 
sol.  AG  :  sol  AK  :  : 
ABCD  :  AMNO. 
But  the  two  parallelopipedons 
AK,AZ,  having  the  same  base 
AMNO,  are  to  each  other  as 
their  altitudes  AE,  AX ;  hence  _ 

we  have 

sol  AK  :  sol  AZ  :  :  AE  :  AX. 
Multiplying  together  the  corresponding  terms  of  these  propor- 
tions, and  omitting  in  the  result  the  common  multiplier  sol  AK ; 
we  shall  have 

sol  AG  :  50/.  AZ  :  :  ABCDxAE  :  AMNO  x  AX. 

Instead  of  the  bases  ABCD  and  AMNO,  put  ABxADand 
AO  X  AM  it  will  give 

50/.AG  :  solAZ  :  :  ABxADxAE  :  AOxAMxAX. 

Hence  any  two  rectangular   parallelopipedons   are  to  each 
other,  &c. 

Scholium.  We  are  consequently  authorized  to  assume,  as 
the  measure  of  a  rectangular  parallelopipedon,  the  product 
of  its,  base  by  its  altitude,  in  other  words,  the  product  of  its 
three  dimensions. 

In  order  to  comprehend  the  nature  of  this  measurement,  it 
is  necessary  to  reflect,  that  the  number  of  linear  units  in  one 
dimension  of  the  base  multiplied  by  the  number  of  Hnear  units 
in  the  other  dimension  of  the  base,  will  give  the  number  of 
superficial  units  in  ihe  base  of  the  parallelopipedon  (Book  IV. 
Prop.  IV.  Sch.).  For  each  unit  in  height  there  are  evidently 
as  many  solid  units  as  there  are  superficial  units  in  the  base. 
Therefore,  the  number  of  superficial  units  in  the  base  multi- 
plied by  the  number  of  linear  units  in  the  altitude,  gives  the 
number  of  solid  units  in  the  parallelopipedon. 

If  the  three  dimensions  of  another  parallelopipedon  are 
valued  according  to  the  same  linear  unit,  and  multiplied  together 
in  the  same  maimer,  the  two  products  will  be  to  each  other  as 


BOOK  VII.  157 

the  solids,  and  will   serve  to  express  their  relative  magni- 
tude. 

The  magnitude  of  a  sohd,  its  volume  or  extent, forms  Vi^hat  is 
called  its  solidity  ;  and  this  word  is  exclusively  employed  to 
designate  the  measure  of  a  solid  :  thus  we  say  the  solidity  of  a 
rectangular  parallelopipedon  is  equal  to  the  product  of  its  base 
by  its  altitude,  or  to  the  product  of  its  three  dimensions. 

As  the  cube  has  all  its  three  dimensions  equal,  if  the  side  is 
1,  the  solidity  will  be  1  x  1  x  1  =  1  :  if  the  side  is  2,  the  solidity 
will  be  2  X  2  x  2  —  8  ;  if  the  side  is  3,  the  solidity  will  be  3  x  3  x 
3  =  27 ;  and  so  on  :  hence,  if  the  sides  of  a  series  of  cubes  are 
to  each  other  as  the  numbers  1,  2,  3,  &c.  the  cubes  themselves 
or  their  solidities  will  be  as  the  numbers  1,  8,  27,  &c.  Hence 
it  is,  that  in  arithmetic,  the  cube  of  a  number  is  the  name  given 
to  a  product  which  results  from  three  factors,  each  equal  to 
this  number. 

If  it  were  proposed  to  find  a  cube  double  of  a  given  cube, 
the  side  of  the  required  cube  would  have  to  be  to  that  of  the 
given  one,  as  the  cube-root  of  2  is  to  unity.  Now,  by  a  geo- 
metrical construction,  it  is  easy  to  find  the  square  root  of  2  ; 
but  the  cube-root  of  it  cannot  be  so  found,  at  least  not  by  the 
simple  operations  of  elementary  geometry,  which  consist  m 
employing  nothing  but  straight  lines,  two  points  of  which  are 
known,  and  circles  whose  centres  and  radii  are  determined. 

Owing  to  this  difficulty  the  problem  of  the  duplication  of 
the  cube  became  celebrated  among  the  ancient  geometers,  as 
well  as  that  of  the  trisection  of  an  angle,  which  is  nearly  of  the 
same  species.  The  solutions  of  which  such  problems  are  sus- 
ceptible, have  however  long  since  been  discovered  ;  and  though 
less  simple  than  the  constructions  of  elementary  geometry,  they 
are  not,  on  that  account,  less  rigorous  or  less  satisfactory. 


PROPOSITION  XIV.    THEOREM. 

The  solidity  of  a  parallelopipedon,  and  generally  of  any  prism, 
is  equal  to  the  product  of  its  hose  by  its  altitude. 

For,  in  the  first  place,  any  parallelopipedon  is  equivalent  to 
a  rectangular  parallelopipedon,  having  the  same  altitude  and 
an  equivalent  base  (Prop.  X.).  Now  the  solidity  of  the  latter 
is  equal  to  its  base  multiplied  by  its  height ;  hence  the  solidity 
of  the  former  is,  in  like  manner,  equal  to  the  product  of  its  base 
by  its  altitude. 

In  the  second  place,  any  triangular  prism  is  half  of  the  par- 
allelopipedon so  constructed  as  to  have  the  same  altitude  and 
a  double  base  (Prop.  VII.).  But  the  solidity  of  the  latter  is  equal 


158 


GEOMETRY. 


to  its  base  multiplied  by  its  altitude  ;  hence  that  of  a  triangular 
prism  is  also  equal  to  the  product  of  its  base,  which  is  half  that 
of  the  parallelopipedon,  multiplied  into  its  altitude. 

In  the  third  place,  any  prism  may  be  divided  into  as  many 
triangular  prisms  of  the  same  altitude,  as  there  are  triangles 
capable  of  being  formed  in  the  polygon  which  constitutes  its 
base.  But  the  solidity  of  each  triangular  prism  is  equal  to  its 
base  multiplied  by  its  altitude  ;  and  since  the  altitude  is  the 
same  for  all,  it  follows  that  the  sum  of  all  the  partial  prisms 
must  be  equal  to  the  sum  of  all  the  partial  triangles,  which  con- 
stitute their  bases,  multiplied  by  the  common  altitude. 

Hence  the  solidity  of  any  polygonal  prism,  is  equal  to  the 
product  of  its  base  by  its  altitude. 

Co7\  Comparing  two  prisms,  which  have  the  same  altitude, 
the  products  of  their  bases  by  their  altitudes  will  be  as  the 
bases  simply  ;  hence  two  prisms  of  the  same  altitude  are  to  each 
other  as  their  bases.  For  a  like  reason,  two  prisms  of  the  same 
base  are  to  each  other  as  their  altitudes.  And  when  neither  their 
bases  nor  their  altitudes  are  equal,  their  solidities  will  be  to 
each  other  as  the  products  of  their  bases  and  altitudes. 


PROPOSITION  XV.     THEOREM. 


TVjo  triangular  pyramids,  having  equivalent  bases  and  equal 
altitudes,  are  equivalent,  or  equal  in  solidity. 


Let  S-ABC,  S-ahc,  be  those  two  pyramids  ;  let  their  equiva- 
lent bases  ABC,  abc,  be  situated  in  the  same  plane,  and  let  AT 
be  their  common  altitude.  If  they  are  not  equivalent,  let  ^-ahe 


BOOK  VII.  159 

be  the  smaller :  and  suppose  Ka  to  be  the  altitude  of  a  prism, 
which  having  ABC  for  its  base,  is  equal  to  their  difference. 

Divide  the  altitude  AT  into  equal  parts  Ax,  xy,  yz,  &c.  each 
less  than  A«,  and  let  k  be  one  of  those  parts  ;  through  the  points 
of  division  pass  planes  parallel  to  the  plane  of  the  bases  ;  the 
icorre spending  sections  formed  by  these  planes  in  the  two  pyra- 
mids will  be  respectively  equivalent,  namely  DEF  to  def,  GHI 
to  ghi,  &c.  (Prop.  III.  Cor.  2.). 

This  being  granted,  upon  the  triangles  ABC,  DEF,  GHI,  &c. 
taken  as  bases,  construct  exterior  prisms  having  for  edges  the 
parts  AD,  DG,  GK,  &c.  of  the  edge  SA  ;  in  like  manner,  on 
bases  dej,  ghi,  klm,  &c.  in  the  second  pyramid,  construct  inte- 
rior prisms,  having  for  edges  the  corresponding  parts  of  Sa. 
It  is  plain  that  the  sum  of  all  the  exterior  prisms  of  the  pyramid 
S-ABC  will  be  greater  than  this  pyramid  ;  and  also  that  the 
sum  of  all  the  interior  prisms  of  the  pyramid  S-abc  will  be  less 
than  this  pyramid.  Hence  the  difference,  between  the  sum  of  all 
the  exterior  prisms  and  the  sum  of  all  the  interior  ones,  must  be 
greater  than  the  difference  between  the  two  pyramids  them- 
selves. 

Now,  beginning  with  the  bases  ABC,  ahc,  the  second  exte- 
rior prism  DEF-G  is  equivalent  to  the  first  interior  prism  def-a, 
because  they  have  the  same  altitude  k,  and  their  bases  DEF, 
def,  are  equivalent ;  for  like  reasons,  the  third  exterior  prism 
GHI-K  and  the  second  interior  prism  ghi-d  are  equivalent ; 
the  fourth  exterior  and  the  third  interior  ;  and  so  on,  to  the  last 
in  each  series.  Hence  all  the  exterior  prisms  of  the  pyramid 
S-ABC,  excepting  the  first  prism  ABC-D,  have  equivalent  cor- 
responding ones  in  the  interior  prisms  of  the  pyramid  S-abc : 
hence  the  prism  ABC-D,  is  the  difference  between  the  sum  of 
all  the  exterior  prisms  of  the  pyramid  S-ABC,  and  the  sum  of 
the  interior  prisms  of  the  pyramid  S-abc.  But  the  difference 
between  these  two  sets  of  prisms  has  already  been  proved  to 
be  greater  than  that  of  the  two  pyramids  ;  which  latter  diffe- 
rence we  supposed  to  be  equal  to  the  prism  a-ABC  :  hence  the 
prism  ABC-D,  must  be  greater  than  the  prism  a-ABC.  But  in 
reality  it  is  less  ;  for  they  have  the  same  base  ABC,  and  the 
altitude  Ax  of  the  first  is  less  than  Aa  the  altitude  of  the  second. 
Hence  the  supposed  inequality  between  the  two  pyramids  can- 
not exist ;  hence  the  two  pyramids  S-ABC,  S-abc,  having  equal 
altitudes  and  equivalent  bases,  are  themselves  equivalent. 


leo  GEOMETRY. 


PROPOSITION  XVI.     THEOREM. 

Ev.ery  triangular  pyramid  is  a  third  part  of  the  triangular  prism 
having  the  same  base  and  the  same  altitude. 

Let  F-ABC  be  a  triangular 
pyramid,  ABC-DEF  a  triangular 
prism  of  the  same  base  and  the 
same  altitude  ;  the  pyramid  will 
be  equal  to  a  third  of  the  prism. 

Cut  off  the  pyramid  F-ABC 
from  the  prism,  by  the  plane 
FAC  ;  there  will  remain  the  solid 
F-ACDE,  which  may  be  consi- 
dered as  a  quadrangular  pyramid, 
whose  vertex  is  F,  and  whose  base 
is  the  parallelogram  ACDE. 
Draw  the  diagonal  CE  ;  and  pass 
the  plane  FCE,  which  will  cut  the  B 

quadrangular  pyramid  into  two  triangular  ones  F-ACE,F-CDE. 
These  two  triangular  pyramids  have  for  their  common  altitude 
the  perpendicular  let  fall  from  F  on  the  plane  ACDE ;  they 
have  equal  bases,  the  triangles  ACE,  CDE  being  halves  of  the 
same  parallelogram  ;  hence  the  two  pyramids  F-ACE,  F-CDE, 
are  equivalent  (Prop.  XV.).  But  the  pyramid  F-CDE  and  the 
pyramid  F-ABC  have  equal  bases  ABC,  DEF;  they  have  also  the 
same  altitude,  namely,  the  distance  between  the  parallel  planes 
ABC,  DEF ;  hence  the  two  pyramids  are  equivalent.  Now  the 
pyramid  F-CDE  has  already  been  proved  equivalent  to  F-ACE  ; 
hence  the  three  pyramids  F-ABC,  F-CDE,  F-ACE,  which 
compose  the  prism  ABC-DEF  are  all  equivalent.  Hence  the 
pyramid  F-ABC  is  the  third  part  of  the  prism  ABC-DEF,  which 
has  the  same  base  and  the  same  altitude. 

Cor.  The  solidity  of  a  triangular  pyramid  is  equal  to  a  third 
part  of  the  product  of  its  base  by  its  altitude. 


PROPOSITION  XVII.    THEOREM. 

The  solidity  of  every  pyramid  is  equal  to  the  base  multiplied  by 
a  third  of  the  altitude,. 


BOOK  VII.  161 

Let  S-ABCDE  be  a  pyramid. 

Pass  the  planes  SEB,  SEC,  through  the 
diagonals  EB,  EC ;  the  polygonal  pyraniid 
S-ABCDE  will  be  divided  into  several  trian- 
gular pyranriids  all  having  the  same  altitude 
SO.  But  each  of  these  pyramids  is  measured 
by  multiplying  its  base  ABE,  BCE,  or  CDE, 
by  the  third  part  of  its  altitude  SO  (Prop.  XVI. 
Cor.) ;  hence  the  sum  of  these  triangular  pyra- 
mids, or  the  polygonal  pyramid  S-ABCDE 
will  be  measured  by  the  sum  of  the  triangles 
ABE,  BCE,  CDE,  or  the  polygon  ABODE,  ^ 

multiplied  by  one  third  of  SO  ;  hence  every  pyramid  Is  mea- 
sured by  a  third  part  of  the  product  of  its  base  by  its  altitude. 

Cor,  1.  Every  pyrarnid  is  the  third  part  of  the  prism  which 
has  the  same  base  and  the  same  altitude. 

Coj\  2.  Two  pyramids  having  the  same  altitude  are  to  each 
other  as  their  bases. 

Cor.  3.  Two  pyramids  havmg  equivalent  bases  are  to  each 
other  as  their  altitudes. 

Cor.  4.  Pyramids  are  to  each  other  as  the  products  of  their 
bases  by  their  altitudes. 

Scholium.  The  solidity  of  any  polyedral  body  may  be  com- 
puted, by  dividing  the  body  into  pyramids ;  and  this  division 
may  be  accomplished  in  various  ways.  One  of  the  simplest 
is  to  make  all  the  planes  of  division  pass  through  the  vertex 
of  one  solid  angle  ;  in  that  case,  there  will  be  formed  as  many 
partial  pyramids  as  the  polyedron  has  faces,  minus  those  faces 
which  form  the  solid  angle  whence  the  planes  of  division 
proceed. 


PROPOSITION  XVIII.    THEOREM. 

If  a  pyramid  be  cut  hy  a  plane  parallel  to  its  base,  the  frustum 
that  remains  when  the  small  pyramid  is  taken  away,  is  equi- 
valent to  the  sum  of  three  pyramids  having  for  their  common 
altitude  the  altitude  of  the  frustum,  and  for  bases  the  lower 
base  of  the  frustum,  the  upper  base,  and  a  mean  proportional 
between  the  two  bases, 

O* 


162 


GEOMETRY 


Let  S-ABCDE  be  a  pyra- 
mid cut  by  the  plane  abcdCf 
parallel  to  its  base;  let  T-FGH 
be  a  triangular  pyramid  hav- 
ing the  same  altitude  and  an 
equivalent  base  with  the  pyra- 
mid S-ABCDE.  The  two 
bases  may  be  regarded  as 
situated  in  the  same  plane  ;  in 
which  case,  the  plane  abed,  if 
produced,  will  form  in  the  triangular  pyramid  a  section  fgk 
situated  at  the  same  distance  above  the  common  plans  of  the 
bases  ;  and  therefore  the  section  j/^/i  will  be  to  the  section  aftc^/e 
as  the  base  FGH  is  to  the  base  ADD  (Prop.  III.),  and  since 
the  bases  are  equivalent,  the  sections  will  be  so  likewise. 
Hence  the  pyramids  S-abcde,  T-fgh  are  equivalent,  for  their 
altitude  is  the  same  and  their  bases  are  equivalent.  The  whole 
pyramids  S-ABCDE,  T-FGH  are  equivalent  for  the  same  rea- 
son ;  hence  the  frustums  ABD-dab,  FGH-hfg  are  equivalent ; 
hence  if  the  proposition  can  be  proved  in  the  single  case  of 
the  frustum  of  a  triangular  pyramid,  it  will  be  true  of  every 
other. 

liCt  FGH-hfg  be  the  frustum  of  a  tri- 
angular pyramid,  having  parallel  bases : 
through  the  three  points  F,  g,  H,  pass 
the  plane  F^H ;  it  will  cut  off  from  the 
frustum  the  triangular  pyramid  g-FGH. 
This  pyramid  has  for  its  base  the  lower 
base  FGH  of  the  frustum  ;  its  altitude 
likewise  is  that  of  the  frustum,  because 
the  vertex  g  lies  in  the  plane  of  the  up- 
per base  fgh. 

This  pyramid  being  cut  off,  there  will 
remain  the  quadrangular  pyramid 
g-f/iKF,  whose  vertex  is  g,  and  base  fJiHF.  Pass  the  plane 
fgH  through  the  three  points  /,  g,  H  ;  it  will  divide  the  quad- 
rangular pyramid  into  two  triangular  pyramids  g-F/H,  g-fhH. 
The  latter  has  for  its  base  the  upper  base  gfh  of  the  frustum ; 
and  for  its  altitude,  the  altitude  of  the  frustum,  because  its  ver- 
tex H  lies  in  the  lower  base.  Thus  we  already  know  two  of 
the  three  pyramids  which  compose  the  frustum. 

It  remains  to  examine  the  third  ^-FfH.  Now,  if  ^K  be 
drawn  parallel  to  /F,  and  if  we  conceive  a  new  pyramid 
K-FfH,  having  K  for  its  vertex  and  FfH  for  its  base,  these 
two  pyramids  will  have  the  same  base  F/"H ;  they  will  also 
have  the  same  altitude,  because  their  vertices  g  and  K  lie  in 
the  line  ^K,  parallel  to  F/,  and  consequently  parallel  to  the 


BOOK  VII.  163 

plane  of  the  base  :  hence  these  pyramids  are  equivalent.  But 
the  pyramid  K-F/H  may  be  regarded  as  having  its  vertex  in 
/,  and  thus  its  altitude  will  be  the  same  as  that  of  the  frufitum : 
as  to  its  base  FKH,  we  are  now  to  show  that  this  is  a  mean 
proportional  between  the  bases  FGH  and  fgh.  Now,  the  tri- 
angles YMYLjfgh,  have  each  an  equal  angle  F=/;  hence 

FHK  :  /^/i  :  :  FKxFH  :  fgxfh  (Book  IV.  Prop.  XXIV.)  ; 
but  because  of  the  parallels,  FK==^,  hence 
FHK  :  /^A  :  :  FH  :  fh. 
We  have  also, 

FHG  :  FHK  :  :  FG  :  FK  or  fg. 
But  the  similar  triangles  FGH,7^/i  give  * 

FG:/^:  :  FH  : /A  ; 
hence, 

FGH  :  FHK  :  :  FHK  :/^A; 
or  the  base  FHK  is  a  mean  proportional  between  the  two 
bases  FGH,  fgh.  Hence  the  frustum  of  a  triangular  pyramid 
is  equivalent  to  three  pyramids  whose  common  altitude  is  that 
of  the  frustum  and  whose  bases  are  the  lower  base  of  the 
frustum,  the  upper  base,  and  a  mean  proportional  between  the 
two  bases. 


PROPOSITION  XIX.    THEOREM. 

Similar  triangular  prisms  are  to  each  other  as  ike  cubes  of  their 
homologous  sides. 

Let  CBD-P,  chd-p,  be  two 
similar  triangular  prisms,  of 
which  BC,  he,  are  homologous 
sides :  then  will  the  prism 
CBD-P  be  to  the  prism  chd-p, 
as  BC^  to  hc^. 

For,  since  the  prisms  are 
similar,  the  planes  which  con- 
tain the  homologous  solid  an-  C  ^  B 
gles  B  and  h,  are  similar,  like  placed,  and  equally  inclined  to 
each  other  (Def.  17.) :  hence  the  solid  angles  B  and  h,  are  equal 
(Book  VI.  Prop.  XXI.  Sch.).  If  these  solid  angles  be  applied 
to  each  other,  the  angle  c/;6?will  coincide  with  CBD,  the  side  ha 
with  B  A,  and  the  prism  chd-p  will  take  the  position  Bcc?-p.  From  A 
draw  AH  perpendicular  to  the  common  base  of  the  prisms :  then 
will  the  plane  BAH  be  perpendicular  to  the  plane  of  the  com- 


164 


GEOMETRY. 


mon  base  (Book  VI.  Prop.  XVI.).  Through  a,  in  the  plane  BAH. 
draw  ah  perpendicular  to 
BH  :  then  will  ah  also  be  per- 
pendicular to  the  base  BDC 
(Book  VI.  Prop.  XVII.)  ;  and 
AH,  ah  will  be  the  altitudes 
of  the  two  prisms. 

Now,  because  of  the  similar 
triangles  ABH,«BA,  an(l  of  the 
similar  parallelograms  AC,  ac, 
we  have 

AH  :  flf/i  :  :  AB  :  ez6  :  :  BC  :  he. 
But  since  the  bases  are  similar,  we  have 
base  BCD  :  base  bed  :  :  BC^  ;  he"^  (Book  IV.  Prop.  XXV.)  ; 
hence, 

base  BCD  :  base  bed  :  :  AH^  :  ah\ 
Multiplying  the  antecedents  by  AH,  and  the  consequents  by 
ahf  and  we  have 

base  BCD  X  AH  :  base  bed  x  ah  :  :  AH^     ah\ 
But  the  solidity  of  a  prism  is  equal  to  the  base  multiplied  by 
the  altitude  (Prop.  XIV.)  ;  hence,  the 

prism  BCD-P  :  prism   bcd-p  :  :  AH^  :  ah^  :  :  BC^  :  bc\ 
or  as  the  cubes  of  any  other  of  their  homologous  sides. 

Cor.  Whatever  be  the  bases  of  similar  prisms,  the  prisms 
will  be  to  each  other  as  the  cubes  of  their  homologous  sides. 

For,  since  the  prisms  are  similar,  their  bases  will  be  similar 
polygons  (Def.  17.) ;  and  these  similar  polygons  may  be  di- 
vided  into  an  equal  number  of  similar  triangles,  similarly  placed 
(Book  IV.  Prop.  XXVI.) :  therefore  the  two  prisms  may  be 
divided  into  an  equal  number  of  triangular  prisms,  having  their 
faces  similar  and  like  placed  ;  and  therefore,  equally  inclined 
(Book  VI.  Prop.  XXI.) ;  hence  the  prisms  will  be  similar.  But 
these  triangular  prisms  will  be  to  each  other  as  the  cubes  of 
their  homologous  sides,  which  sides  being  proportional,  the 
sums  of  the  triangular  prisms,  that  is,  the  polygonal  prisms,  will 
be  to  each  other  as  the  cubes  of  their  homologous  sides. 


I 


PROPOSITION  XX.    THEOREM. 

Two  similar  pyramids  are  to  each  other  as  the  cubes  of  their 
homologous  sides. 


BOOK  VII.  165 

For,  since  the  pyramids  are  similar,  the  soUd 
angles  at  the  vertices  will  be  contained  by  the 
same  number  of  similar  planes,  like  placed, 
and  equally  inchned  to  each  other  (Def.  17.). 
Hence,  the  solid  angles  at  the  vertices  may 
be  made  to  coincide,  or  the  tv^^o  pyramids 
may  be  so  placed  as  to  have  the  solid  angle 
S  common. 

In  that  position,  the  bases  ABCDE,  abcde, 
vf'iW  be  parallel ;  because,  since  the  homolo- 
gous faces  are  similar,  the  angle  S<z&  is  equal 
to  SAB,  and  S6c  to  SBC ;  hence  the  plane 
ABC  is  parallel  to  the  plane  ahc  (Book  VI.  Prop.  XIII.).  This 
being  proved,  let  SO  be  the  perpendicular  drawn  from  the 
vertex  S  to  the  plane  ABC,  and  o  the  point  where  this  perpen- 
dicular meets  the  plane  ahc :  from  what  has  already  been 
shown,  we  shall  have 

SO  :  So  :  :  SA  :  Sa  :  :  AB  :    ah  (Prop.  III.) ; 
and  consequently, 

iSO  :  iSo  :  :  AB  :  ah. 
But  the  bases  ABCDE,  abcde,  being  similar  figures,  we  have 
ABCDE  :  abode  :  :  AB^  :  aU'  (Book  IV.  Prop.  XXVIL). 
Multiply  the  corresponding  terms  of  these  two  proportions ; 
there  results  the  proportion, 

ABCDE  xiSO  :  ahcdex^So  :  :  AB^  :  ah\ 
Now  ABCDE  X  iSO  is  the  solidity  of  the  pyramid  S-ABCDE, 
and  abcdexjSo  is  that  of  the  pyramid  S-abcde  (Prop.  XVII.)  ; 
hence  two  similar  pyramids  are  to  each  other  as  the  cubes  of 
their  homologous  sides. 

General  Scholium. 

The  chief  propositions  of  this  Book  relating  to  the  solidity  ol 
polyedrons,  may  be  exhibited  in  algebraical  terms,  and  so 
recapitulated  in  the  briefest  manner  possible. 

Let  B  represent  the  base  of  a  prism  ;  H  its  altitude  :  the 
solidity  of  the  prism  will  be  B  x  H,  or  BH. 

Let  B  represent  the  base  of  a  pyramid ;  H  its  altitude :  the 
solidity  of  the  pyramid  will  be  B  x  ^H,  or  H  x  ^B,  or  ^BH. 

Let  H  represent  the  altitude  of  the  frustum  of  a  pyramid, 
having  parallel  bases  A  and  B  ;  VAB  will  be  the  mean  pro- 
portional between  those  bases ;  and  the  solidity  of  the  frustum 
willbeiHx(A  +  B+VAB). 

In  fine,  let  P  and  p  represent  the  solidities  of  two  similar 
prisms  or  pyramids ;  A  and  a,  two  homologous  edges  :  then  we 
shall  have 

P  :  p  ;  :  A3  :  a\ 


166 


GEOMETRY. 


BOOK  VIII. 


THE  THREE  ROUND  BODIES. 


Definitions. 


E 


M 


:n: 


\TI> 


^:::^P^^ 


Li 


K 


G 


1.  A  cylinder  is  the  solid  generated  by  the  revolution  oif  a 
rectangle  ABCD,  conceived  to  turn  about  the  immoveable 
side  AB. 

In  this  movement,  the  sides  AD,  BC,  con-         ^ 
tinuing  always  perpendicular  to  AB,  describe 
equal  circles  DHP,  CGQ,  which  are  called 
the  bases  of  the  cylinder,  the  side  CD  at  the 
same  time  describing  the  convex  surface. 

The  immoveable  line  AB  is  called  the  axis 
of  the  cylinder. 

Every  section  KLM,  made  in  the  cylinder, 
at  right  angles  to  the  axis,  is  a  circle  equal  to 
either  of  the  bases  ;  for,  whilst  the  rectangle 
ABCD  turns  about  AB,  the  line  KI,  perpen- 
dicular to  AB,  describes  a  circle,  equal  to  the  base,  and  this 
circle  is  nothing  else  than  the  section  made  perpendicular  to 
the  axis  at  the  point  I. 

Every  section  PQG,  made  through  the  axis,  is  a  rectangle 
double  of  the  generating  rectangle  ABCD. 

2.  A  cone  is  the  solid  generated  by  the  revolution  of  a  right- 
angled  triangle  SAB,  conceived  to  turn  about  the  immoveable 
side  SA. 

In  this  movement,  the  side  AB  describes 
a  circle  BDCE,  named  the  base  of  the  cone  ; 
the  hypothenuse  SB  describes  the  convex 
surface  of  the  cone. 

The  point  S  is  named  the  vertex  of  the 
cone,  SA  the  axis  or  the  altitude,  and  SB 
the  side  or  the  apothem. 

Every  section  HKFI,  at  right  angles  to 
the  axis,  is  a  circle  ;  every  section  SDE, 
through  the  axis,  is  an  isosceles  triangle, 
double  of  the  generating  triangle  SAB. 

3.  If  from  the  cone  S-CDB,  the  cone  S-FKH  be  cut  off  by 
a  plane  parallel  to  the  base,  the  remaining  sohd  CBHF  is  called 
a  truncated  cone,  or  the  frustum  of  a  cone. 


BOOK  VIII. 


167 


We  may  conceive  it  to  be  generated  by  the  revolution  of  a 
trapezoid  ABHG,  whose  angles  A  and  G  are  right  angles,  about 
the  side  AG.  The  immoveable  line  AG  is  called  the  axis  or 
altitude  of  the  frustum^  the  circles  BDC,  HEK,  are  its  bases,  and 
BH  is  its  side. 

4.  Two  cylinders,  or  two  cones,  are  similar,  when  their 
axes  are  to  each  other  as  the  diameters  of  their  bases. 

5.  If  in  the  circle  ACD,  which  forms  the 
base  of  a  cylinder,  a  polygon  ABCDE  be 
inscribed,  a  right  prism,  constructed  on  this 
base  ABCDE,  and  equal  hi  altitude  to  the 
cylinder,  is  said  to  be  inscribed  in  the  cylin- 
der, or  the  cylinder  to  be  circumscribed 
about  the  prism. 

The  edges  AF,  BG,  CH,  &c.  of  the  prism, 
being  perpendicular  to  the  plane  of  the  base, 
are  evidently  included  in  the  convex  sur- 
face of  the  cylinder  ;  hence  the  prism  and 
the  cylinder  touch  one  another  along  these 
edges. 

6.  In  like  manner,  if  ABCD  is  a  poly- 
gon, circumscribed  about  the  base  of  a 
cylinder,  a  right  prism,  constructed  on  this 
base  ABCD,  and  equal  in  altitude  to  the 
cylinder,  is  said  to  be  circumscribed  about 
the  cylinder,  or  the  cylinder  to  be  inscribed 
in  the  prism. 

Let  M,  N,  &c.  be  the  points  of  contact 
in  the  sides  AB,  BC,  &c.  ;  and  through  the 
points  M,N,&c.  let  MX,  NY,  &c.  be  drawn  Ak 
perpendicular  to  the  plane  of  the  base :       x^ 
these  perpendiculars  will  evidently  lie  both 
in  the  surface  of  the  cylinder,  and  in  that  13 

of  the  circumscribed  prism ;  hence  they  will  be  their  lines  of 
contact. 

7.  If  in  the  circle  ABCDE,  which  forms 
the  base  of  a  cone,  any  polygon  ABCDE 
be  inscribed,  and  from  the  vertices  A,  B, 
C,  D,  E,  lines  be  drawn  to  S,  the  vertex 
of  the  cone,  these  lines  may  be  regarded 
as  the  sides  of  a  pyramid  whose  base  is 
the  polygon  ABCDE  and  vertex  S.  The 
sides  of  this  pyramid  are  in  the  convex 
surface  of  the  cone,  and  the  pyramid  is 
said  to  be  inscribed  in  the  cone. 


168 


GEOMETRY. 


8.  The  sphere  is  a  solid  terminated  by  a  curved  surface,  all 
the  points  of  which  are  equally  distant  from  a  point  within, 
called  the   centre.  "- 

The  sphere  may  be  con- 
ceived to  be  generated  by  the 
revolution  of  a  semicircle 
DAE  about  its  diameter  DE : 
for  the  surface  described  in 
this  movement,  by  the  curve 
DAE,  will  have  all  its  points 
equally  distant  from  its  cen- 
tre C. 

9.  Whilst  the  semicircle 
DAE  revolving  round  its  di- 
ameter DE,  describes  the 
sphere ;  any  circular  sector, 
as  pCF  or  FCH,  describes  a 
solid,  which  is  named  a  spherical  sector. 

10.  The  radius  of  a  sphere  is  a  straight  line  drawn  from  the 
centre  to  any  point  of  the  surface  ;  the  diameter  or  axis  is  a . 
line  passing  through  this  centre,  and  terminated  on  both  sides , 
by  the  surface. 

All  the  radii  of  a  sphere  are  equal ;  all  the  diameters  are 
equal,  and  each  double  of  the  radius. 

11.  It  will  be  shown  (Prop.  VII.)  that  every  section  of  the 
sphere,  made  by  a  plane,  is  a  circle  :  this  granted,  a  great  cir-^ 
cle  is  a  section  which  passes  through  the  centre  ;  a  small  circle^ 
is  one  which  does  not  pass  through  the  centre. 

12.  A  plane  is  tangent  to  a  sphere,  when  their  surfaces  have 
but  one  point  in  common: 

13.  A  zone  is  a  portion  of  the  surface  of  the  sphere  included 
between  two  parallel  planes,  which  form  its  bases.  One  of 
these  planes  may  be  tangent  to  the  sphere  ;  in  which  case,  the 
zone  has  only  a  single  base. 

14.  A  spherical  segment  is  the  portion  of  the  solid  sphere, 
included  between  two  parallel  planes  which  form  its  bases. 
One  of  these  planes  may  be  tangent  to  the  sphere  ;  in  which 
case,  the  segment  has  only  a  single  base. 

15.  The  altitude  of  a  zone  or  of  a  segment  is  the  distance 
between  the  tw^o  parallel  planes,  which  form  the  bases  of  the 
zone  or  segment. 

Note.  The  Cylinder,  the  Cone,  and  the  Sphere,  are  the 
three  round  bodies  treated  of  in  the  Elements  of  Geometry. 


BOOK  VIII. 


1G9 


PROPOSITION  I.    THEOREM. 


ne  convex  surface  of  a  cylinder  is  equal  to  the  circumference  of 
its  base  multiplied  by  its  altitude. 

Let  CA  be  the  radius  of  the 
given  cylinder's  base,  and  H  its 
altitude  :  the  circumference 
whose  radius  is  CA  being  rep- 
resented by  circ.  CA,  we  are  to 
show  that  the  convex  surface  of 
the  cyhnder  is  equal  to  circ.  CA 
xH. 

Inscribe  in  the  circle  any 
regular  polygon,  BDEFGA,  and 
construct  on  this  polygon  a  right 
prism  having  its  altitude  equal  to  H,  the  altitude  of  the  cylin- 
der :  this  prism  will  be  inscribed  in  the  cylinder.  The  convex 
surface  of  the  prism  is  equal  to  the  perimeter  of  the  polygon, 
multiphed  by  the  altitude  H  (Book  VII.  Prop.  I.).  Let  now 
the.  arcs  which  subtend  the  sides  of  the  polygon  be  continually 
bisected,  and  the  number  of  sides  of  the  polygon  indefinitely 
increased  :  the  perimeter  of  the  polygon  will  then  become  equal 
to  circ.  CA  (Book  V.  Prop.  VIII.  Cor.  2.),  and  the  convex  sur- 
face of  the  prism  will  coincide  with  the  convex  surface  of  the 
cylinder.  But  the  convex  surface  of  the  prism  is  equal  to  the 
perimeter  of  its  base  multiplied  by  H,  whatever  be  the  number 
of  sides  :  hence,  the  convex  surface  of  the  cylinder  is  equal  to 
the  circumference  of  its  base  multiplied  by  its  altitude. 


PROPOSITION  II.    THEOREM. 


The  solidity  of  a  cylinder  is  equal  to  the  product  of  its  base  by  its 

altitude. 


I'TO  GEOMETRY. 

Let  CA  be  the  radius  of  the 
base  of  the  cylinder,  and  11 
ihe  altitude.  Let  the  circle 
whose  radius  is  CA  be  repre- 
sented by  area  CA,  it  is  to  be 
proved  that  the  solidity  of  the 
cylinder  is  equal  to  area  CA  x  Hi 
Inscribe  in  the  circle  any  regu- 
lar polygon  BDEFGA,  and  con- 
struct on  this  polygon  a  right 
prism  having  its  altitude  equal 
to  H,  the  altitude  of  the  cylinder :  this  prism  will  be  inscribed 
in  the  cylinder.  The  solidity  of  the  prism  will  be  equal  to  the 
area  of  the  polygon  multiplied  by  the  altitude  H  (Book  VIL 
Prop.  XIV.).  Let  now  the  number  of  sides  of  the  polygon  be 
indefinitely  increased  :  the  solidity  of  the  new  prism  will  still 
be  equal  to  its  base  multiplied  by  its  altitude. 

But  when  the  number  of  sides  of  the  polygon  is  indefinitely 
increased,  its  area  becomes  equal  to  the  area  CA,  and  its  pe- 
rimeter coincides  with  circ.  CA  (Book  V.  Prop.  VUL  Cor.  1. 
&  2.)  ;  the  inscribed  prism  then  coincides  with  the  cylinder, 
since  their  altitudes  are  equal,  and  their  convex  surfaces  per- 
pendicular to  the  common  base  :  hence  the  two  solids  will  be 
equal ;  therefore  the  solidity  of  a  cylinder  is  equal  to  the  product 
of  its  base  by  its  altitude. 

Cor.  L  Cylinders  of  the  same  altitude  are  to  each  other  as 
their  bases  ;  and  cylinders  of  the  same  base  are  to  each  other 
as  their  altitudes. 

-i;.i4^-  -:v-.  ■■■■  '  '     • 
^•-'•Coi\  2.    Similar  cylindo  -s  are  to  each  other  as  the  cubes  of 

their  altitudes,  or  as  the  cuoes  of  the  diameters  of  their  bases. 

For  the.  bases  are  as  the  squares  of  their  diameters  ;  and  the 
,  cylinders  behig  similar,  the  diameters  of  their  bases  are  to 
.  each  other  as  the  altitudes  (Def.  4.)  ;    hence  the  bases  are 

as  the  squares  of  the  altitudes  ;  hence  the  bases,  multiplied 

by  the  altitudes,  or  the  cylinders  themselves,  are  as  the  cubes 

of  the  altitudes. 

Scholium,  Let  R  be  the  radius  of  a  cylinder's  base  ;  H  the 
altitude  :  the  surface  of  the  base  will  be  tt.W  (Book  V.  Prop. 
XII.  Cor.  2.)  ;  and  the  solidity  of  the  cylinder  will  be  jiR-xH 
or^r.Rs.H. 


BOOK  VIII. 


171 


PROPOSITION  III.    THEOREM. 


The  convex  surface  of  a  cone  is  equal  to  the  circumference  of  its 
'■  haselmultipliedhy  half  its  side. 


Let  the  circle  ABCD  be  the 
base  of  a  cone,  S  the  vertex, 
SO  the  altitude,  and  SA  the 
side  :  then  will  its  convex  sur- 
face be  equal  to  circ.  OA  x  ^S  A. 

For,  inscribe  in  the  base  of 
the  cone  any  regular  polygon 
ABCD,  and  on  this  polygon  as 
a  base  conceive  a  pyramid  to 
be  constructed  having  S  for  its 
vertex :  this  pyramid  will  be  a 
regular  pyramid,  and  will  be  inscribed  in  the  cone. 

From  S,  drav/  SG  perpendicular  to  one  of  the  sides  of  the 
polygon.  The  convex  surface  of  the  inscribed  pyramid  is  equal 
to  the  perimeter  of  the  polygon  which  forms  its  base,  multiplied 
by  half  the  slant  height  SG  (Book  VII.  Prop.  IV.).  Let  now 
the  number  of  sides  of  the  inscribed  polygon  be  indefinitely 
increased;  the  perimeter  of  the  inscribed  polygon  will  then 
become  equal  to  circ,  OA,  the  slant  height  SG  will  become 
equal  to  the  side  SA  of  the  cone,  and  the  convex  surfape  of 
the  pyramid  to  the  convex  surface  of  the  cone.  But  wfe-atever 
be  the  number  of  sides  of  the  polygon  which  forms 'the  base, 
the  convex  surface  of  the  pyramid  is  equal  to  the  perimeter  of 
the  base  multiplied  by  half  the  slant  height:  hence  the.  convex 
.surface  of  a  cone  is  equal  to  the  circumference  of  the  base 
multiplied  by  half  the  side. 

Scholium.  Let  L  be  the  side  of  a  cone,  R  the  radius. of  its 
base  ;  the  circumference  of  this  base  will  be  27r.R,  and  the  sur- 
face of  the  cone  will  be  2iR  x^L,  or  jiRL. 


y?{ 


PROPOSITION  IV.    THEOREM. 


The  convex  surface  of  the  frustum  of  a  cone  is  equal  to  its  side 
multiplied  by  half  the  sum  of  the  circumferences  of  its  two 
bases. 


172 


GEOMETRY 


Lei  BTA-DE  be  a  frustum  of  a 
cone :  then  will  its  convex  surface  be 
equal  to  AD  x  ^circ.OA  +  circ.CU^ 

.  For,  inscribe  in  the  bases  of  the 
frustums  two  regular  polygons  of  the 
same  number  of  sides,  and  having 
their  homologous  sides  parallel,  each 
to  each.  The  lines  joining  the  ver- 
tices of  the  homologous  angles  may 
be  regarded  as  the  edges  of  the  frus- 
tum of  a  regular  pyramid  inscribed 
in  the  frustum  of  the  cone.  The  con- 
vex surface  of  the  frustum  of  the 
pyramid  is  equal  to  half  the  sum  of  the  perimeters  of  its  bases 
multiplied  by  the  slant  height  fh  (Book  VII.  Prop.  IV.  Cor.). 
Let  now  the  number  of  sides  of  the  inscribed  polygons  be 
^  indefinitely  increased :  the  perimeters  of  the  polygons  will  be- 
come equal  to  the  circumferences  BI  A,  EGD  ;  the  slant  height 
fh  will  become  equal  to  the  side  AD  or  BE,  and  the  surfaces 
of  the  two  frustums  will  coincide  and  become  the  same  surface. 
But  the  convex  surface  of  the  frustum  of  the  pyramid  will 
still  be  equal  to  half  the  sum  of  the  perimeters  of  the  upper 
and  low^er  bases  multiplied  by  the  slant  height :  hence  the  sur- 
face of  the  frustum  of  a  cone  is  equal  to  its  side  multiphed  by 
half  the  sum  of  the  circumferences  of  its  two  bases. 


Cor,  Through/,  the  middle  point  of  AD,  draw  /KL  paral- 
*  lel  to  AB,  and  /i,  Dd,  parallel  to  CO.  Then,  since  A/,  11),  are 
'^  equal,  Ai,  id,  will  also  be  equal  (Book  IV.  Prop.  XV.  Cor.  2.) : 
hence,  K/  is  equal  to  ^(OA  +  CD).  But  since  the  circumfe- 
rences of  circles  are  to  each  other  as  their  radii  (Book  V. 
Prop.  XL),  the  circ.  Kl=l(circ.  0A  + circ.  CD) ;  therefore,  the 
convex  surface  of  a  frustum  of  a  cone  is  equal  to  its  side  multi- 
plied hy  the  circumference  of  a  section  at  equal  distances  from 
the  two  bases.  ,  , 


Scholium.  If  a  line  AD,  lying  wholly  on  one  side  of  the  line 
OC,  and  in  the  same  plane,  make  a  revolution  'around  OC, 
tlie  surface,  described  by  AD  will  have  for  its  measure  ADx 
/circ.AQ  + arc. DC\  ^^  ^j^  ^  ^.^^  ^j^.  '^^^  jj^^^  ^^  ^^^  ^^ 

bemg  perpendiculars,  let  fall  from  the  extremities  and  from 
the  middle  point  of  AD,  on  the  axis  OC. 

For,  if  AD  and  OC  are  produced  till  they  meet  in  S,  the 
surface  described  by  AD  is  evidently  the  frustum  of  a  cone 


BOOK  VIII.  173 

• 
having  AO  and  DC  for  the  radii  of  its  bases,  the  vertex  of 
the  whole  cone  being  S.  Hence  this  surface  will  be  measured 
as  we  have  said. 

This  measure  will  always  hold  good,  even  when  the  point 
P  falls  on  S,  and  thus  forms  a  whole  cone ;  and  also  when  the 
line  AD  is  parallel  to  the  axis,  and  thus  forms  a  cylinder.  In 
the  first  case  DC  would  be  nothing ;  in  the  second,  DC  would 
be  equal  to  AO  and  to  IK, 


PROPOSITION  V.    THEOREM. 

The  solidity  of  a  cone  is  equal  to  its  base  multiplied  hy  a  third  of 
its  altitude. 

Let  SO  be  the  altitude  of  a  cone, 
OA  the  radius  of  its  base,  and  let 
the  area  of  the  base  be  designated 
by  area  OA  :  it  is  to  be  proved  that 
the  solidity  of  the  cone  is  equal  to 
area.  OAx^SO. 

Inscribe  in  the  base  of  the  cone 
any  regular  polygon  ABDEF,  and 
join  the  vertices  A,  B,  C,  <fec.  with 
the  vertex  S  of  the  cone  :  then  will 
there  be  inscribed  in  the  cone  a 
regular  pyramid  having  the  same  vertex  as  the  cone,  and  hav- 
ing for  its  base  the  polygon  ABDEF.  The  solidity  of  this 
pyramid  is  equal  to  its  base  multiplied  by  one  third  of  its  ahi- 
tude  (Book  VII.  Prop.  XVII.).  Let  now  the  number  of  sides 
of  the  polygon  be  indefinitely  increased  :  the  polygon  will  then 
become  equal  to  the  circle,  and  the  pyramid  and  cone  will 
coincide  and  become  equal.  But  the  solidity  of  the  pyramid 
is  equal  to  its  base  multiplied  by  one  third  of  its  altitude,  what- 
ever be  the  number  of  sides  of  the  polygon  which  forms  its 
base :  hence  the  solidity  of  the  cone  is  equal  to  its  base  multi- 
plied by  a  third  of  its  altitude. 

Cor.  A  cone  is  the  third  of  a  cylinder  having  the  same  base 
and  the  same  altitude ;  whence  it  follows, 

1.  That  cones  of  equal  altitudes  are  to  each  other  as  their 
bases  ; 

2.  That  cones  of  equal  bases  are  to  each  other  as  their 
altitudes ; 

3.  That  similar  cones  are  as  the  cubes  of  the  diameters  of 
their  bases,  or  as  the  cubes  of  their  altitudes. 

P* 


174  GEOMETRY. 

Cor.  2.  The  solidity  of  a  cone  is  equivalent  to  the  solidity  of 
a  pyramid  having  an  equivalent  base  and  the  same  altitude 
(Book  VII.  Prop.  XVIL). 

Scholium.  Let  R  be  the. radius  of  a  cone's  base,  H  its  alti- 
tude ;  the  solidity  of  the  cone  will  be  nW  x  ^H,  or  ^^iR^H. 


PROPOSITION  VI.    THEOREM 

The  solidity  of  the  frustum  of  a  cone  is  equal  to  the  sum  of  the 
solidities  of  three  cones  whose  common  altitude  is  the  altitude 
of  the  frustum,  and  whose  bases  are,  the  upper  base  of  the  frus- 
tum, the  lower  base  of  the  frustum,  and  a  mean  proportional 
between  them. 

Let  AEB-CD  be  the  frustum  of  a 
cone,  and  OP  its  altitude  ;  tlien  will  its 
solidity  be  equal  to 

^^  X  OP  X  (A02+DP2+ AO  X  DP). 
For,  inscribe  in  the  lower  and-  upper 
basea  two  regular  polygons  having  the 
same  number  of  sides,  and  having  their 
homologous  sides  parallel,  each  to  each. 
Join  the  vertices  of  the  homologous 
angles  and  there  will  then  be  inscribed 
in  the  frustum  of  the  cone,  the  frustum 
of  a  regular  pyramid.     The  sohdity  of 

the  frustum  of  the  pyramid  is  equivalent  to  three  pyramids 
having  the  common  altitude  of  the  frustum,  and  for  bases,  the 
lower  base  of  the  frustum,  the  upper  base  of  the  frustum,  and 
a  mean  proportional  between  them  (Book  VIL  Prop.  XVIII.). 

Let  now,  the  number  of  sides  of  the  inscribed  polygons  be 
indefinitely  increased  :  the  bases  of  the  frustum  of  the  pyramid 
■will  then  coincide  with  the  bases  of  the  frustum  of  the  cone, 
and  the  two  frustums  will  coincide  and  become  the  same  solid. 
Since  the  area  of  a  circle  is  equal  to  R-.tt  (Book  V.  Prop.  XII. 
Cor.  2.),  the  expression  for  the  solidities  of  the  frustum  will 
become 

for  the  first  pyramid        -\0P  x  OA^rr. 
for  the  second  iOP  x  PD^.^r 

for  the  third  J  OP  x  AO  x  PD.Tt ;  since 

AO  x  PD.Ti  is  a  mean  proportional  between  OA^.^r  and  PD^.rr. 
Hence  the  solidity  of  the  frustum  of  the  cone  is  measured  by 
4«0P  x  (OAH PDH  AO  X  PD). 


BOOK  VIII.  175 

PROPOSITION  VII.    THEOREM. 
Every  section  of  a  sphere,  made  by  a  plane,  is  a  cir.c!e. 

Let  AMB  be  a  section,  made  by  a 
plane,  in  the  sphere  whose  centre  is  C. 
From  the  point  C,  draw  CO  perpen- 
dicular to  the  plane  AMB  ;  and  diffe- 
rent lines  CM,  CM,  to  different  points 
of  the  curve  AMB,  which  terminates 
the  section. 

The  oblique  lines  CM,  CM,  CA,  are 
equal,  being  radii  of  the  sphere  ;  hence 
they  are  equally  distant  from  the  perpendicular  CO  (Book  VI. 
Prop.  V.  Cor.) ;  therefore  all  the  lines  OM,  OM,  OB,  are  equal ; 
consequently  the  section  AMB  is  a  circle,  whose  centre  is  O. 

Cor  1.  If  the  section  passes  through  the  centre  of  the  sphere, 
its  radius  will  be  the  radius  of  the  sphere;  hence  all  great 
circles  are  equal. 

Co7\  2.  Two  great  circles  always  bisect  each  other ;  for 
their  common  intersection,  passing  through  the  centre,  is  a 
diameter. 

Cor.  3.  Every  great  circle  divides  the  sphere  and  its  surface 
into  two  equal  parts :  for,  if  the  two  hemispheres  were  sepa- 
rated and  afterwards  placed  on  the  common  base,  with  their 
convexities  turned  the  same  way,  the  two  surfaces  would 
exactly  coincide,  no  point  of  the  one  being  nearer  the  centre 
than  any  point  of  the  other. 

Cor.  4.  The  centre  of  a  small  circle,  and  that  of  the  sphere, 
are  in  the  same  straight  line,  perpendicular  to  the  plane  of  the 
small  circle. 

Cor.  5.  Small  circles  are  the  less  the  further  they  lie  from 
the  centre  of  the  sphere ;  for  the  greater  CO  is,  the  less  is  the 
chord  AB,  the  diameter  of  the  small  circle  AMB. 

Cor.  6.  An  arc  of  a  great  circle  may  always  be  made  to  pass 
through  any  two  given  points  of  the  surface  of  the  sphere  ;  for 
the  two  given  points,  and  the  centre  of  the  sphere  make  three 
points  which  determine  the  position  of  a  plane.  But  if  the 
two  given  points  were  at  the  extremities  of  a  diameter,  these 
two  points  and  the  centre  would  then  lie  in  one  straight  line, 

\  and  an  infinite  number  of  great  circles  might  be  made  to  pass 

s  through  the  two  given  points. 


176 


GEOMETRY. 


PROPOSITION  VIII.    THEOREM. 


Evejy  plane  perpendicular  to  a  radius  at  its  extremity  is  tangent 
to  the.  sphere. 

Let  FAG  be  a  plane  perpendicular 
to  the  radius  OA.  at  its  extremity  A. 
Any  point  M  in  this  plane  being  as- 
sumed, and  OM,  AM,  being  drawn, 
the  angle  0AM  will  be  a  right  angle, 
and  hence  the  distance  OM  will  be 
greater  than  OA.  Hence  the  point 
M  lies  without  the  sphere  ;  and  as  the 
same  can  be  shown  for  every  other 
point  of  the  plane  FAG,  this  plane  can 
have  no  point  but  A  common  to  it  and  the  surface  of  the  sphere ; 
hence  it  is  a  tangent  plane  (Def.  12.) 

Scholium.  In  the  same  way  it  may  be  shown,  that  two 
spheres  have  but  one  point  in  common,  and  therefore  touch 
each  other,  when  the  distance  between  their  centres  is  equal  to 
the  sum,  or  the  difference  of  their  radii ;  in  which  case,  the 
centres  and  the  point  of  contact  lie  in  the  same  straight  line 


PROPOSITION  IX.    LEMMA. 


If  a  regular  semi-polygon  he  rev^olved  about  a  line  passing 
through  the  centre  and  the  vertices  of  two  opposite  angles,  the 
surface  described  by  its  perimeter  will  be  equal  to  the  axis  mul- 
tiplied by  the  circumference  of  the  inscribed  circle. 


Let  the  regular  semi-polygon  ABCDEF, 
be  revolved  about  the  line  AF  as  an  axis  : 
then  will  the  surface  described  by  its  pe- 
rimeter be  equal  to  AF  multiplied  by  the 
circumference  of  the  inscribed  circle. 

From  E  and  D,  the  extremities  of  one  of 
the  equal  sides,  let  fall  the  perpendiculars 
EH,  DI,  on  the  axis  AF,  and  from  the  cen- 
tre O  draw  ON  perpendicular  to  the  side 
DE :  ON  will  be  the  radius  of  the  inscribed 
circle  (Book  V.  Prop.  H.).  Now,  the  sur- 
face described  in  the  revolution  by  any  one 
side  of  the  regular  -polygon,  as  DE,  has 


BOOK  VIII. 


177 


been  shown  to  be  equal  to  DE  x  circ.  NM  (Prop.  IV.  Sch.). 
Biit  since  the  triangles  EDK,  ONM,  are   similar  (Book  IV 
Prop.  XXL),  ED  :  KK  or  HI  :  :  ON  :  NM,  or  as  circ.  ON  . 
circ.  NM ;  hence 

ED  X  circ.  NM = HI  x  circ.  ON  ; 
and  since  the  same  may  be  shown  for  each  of  the  other  sides, 
it  is  plain  that  the  surface  described  by  the  entire  perimeter  i$ 
equal  to 

(FH  +  HH-IP  +  PQ+QA)xaVc  ON=AFxcirc.  ON. 

Cor.  The  surface  described  by  any  portion  of  the  perime- 
ter,  as  EDC,  is  equal  to  the  distance  between  the  two  perpen 
diculars  let  fall  from  its  extremities  on  the  axis,  multiplied  by 
the  circumference  of  the  inscribed  circle.  For,  the  surfac( 
described  by  DE  is  equal  to  HI  x  circ.  ON,  and  the  surface 
described  by  DC  is  equal  to  IPx  circ.  ON  :  hence  the  surface 
described  by  ED  +  DC,  is  equal  to  (HI  +  IP)  x  arc.  ON,  o] 
equal  to  HP  x  circ.  ON. 


PROPOSITION  X.     THEOREM. 

The  surface  of  a  sphere  is  equal  to  the  product  of  its  diameter  5y 
the  circumference  of  a  great  circle. 

Let  ABODE  be  a  semicircle.  Inscribe  in 
it  any  regular  semi-polygon,  and  from  the 
centre  O  draw  OF  perpendicular  to  one  of 
the  sides. 

Let  the  semicircle  and  the  semi-polygon 
be  revolved  about  the  axis  AE :  the  semi- 
circumference  ABODE  will  describe  the 
surface  of  a  sphere  (Def.  8.) ;  and  the  pe- 
rimeter of  the  semi-polygon  will  describe 
a  surface  which  has  for  its  measure  AE  x 
circ.  OF  (Prop.  IX.),  and  this  will  be  true 
whatever  be  the  number  of  sides  of  the  po- 
lygon. But  if  the  number  of  sides  of  the  polygon  be  indefi- 
nitely increased,  its  perimeter  will  coincide  with  the  circumfe- 
rence ABODE,  the  perpendicular  OF  will  become  equal  to 
OE,  and  the  surface  described  by  the  perimeter  of  the  semi- 
polygon  will  then  be  the  same  as  that  described  by  the  semi- 
circumference  ABODE.  Hence  the  surface  of  the  sphere  is 
equal  to  AE  x  circ,  OE. 

Cor.  Since  the  area  of  a  great  circle  is  equal  to  the  product 
of  its  circumference  by  half  the  radius,  or  one  fourth  of  the 


178 


GEOMETRY. 


uiameter  (Book  V.  Prop.  XII.),  it  follows  that  the  surface  of  a 
siihere  is  equal  to  four  of  its  great  circles :  that  is,  equal  to 
4.^.0A2  (Book  V.  Prop.  XII.  Cor.  2.). 

Scholium  1.  The  surface  of  a  zone  is  equal  to  its  altitude  mul- 
tiplied by  the  circumference  of  a  great  circle. 

For,  the  surface  described  by  any  portion 
of  the  perimeter  of  the  inscribed  polygon,  as 
BC  +  CD,  is  equal  to  EH  xcfrc.  OF  (Prop. 
IX.  Cor.).  But  when  the  number  of  sides 
of  the  polygon  is  indefinitely  increased,  BC 
f  CD,  becomes  the  arc  BCD,  OF  becomes 
equal  to  OA,  and  the  surface  described  by 
BC  +  CD,  becomes  the  surface  of  the  zone 
described  by  the  arc  BCD :  hence  the  sur- 
*kce  of  the  zone  is  equal  to  EH  x  circ.  OA. 

Scholium  2.  When  the  zone  has  but  one 
base,  as  the  zone  described  by  the  arc  ABCD,  its  surface  will 
still  be  equal  to  the  altitude  AE  multiplied  by  the  circumference 
of  a  great  circle. 

Scholium  3.  Two  zones,  taken  in  the  same  sphere  or  in  equal 
spheres,  are  to  each  other  as  their  altitudes  ;  and  any  zone  is  to 
the  surface  of  the  sphere  as  the  altitude  of  the  zone  is  to  the 
diameter  of  the  sphere. 


PROPOSITION  XL    LEMMA. 


If  a  triangle  and  a  rectangle,  having  the  same  base  and  the  same 
altitude,  tujm  together  about  the  common  base,the  solid  described 
by  the  triangle  will  be  a  third  of  the  cylinder  described  by  the 
rectangle.  *^' 

Let  ACB  be  the  triangle,  and  BE  the  rectangle. 

On  the  axis,  let  fall  the  perpcn-  ^ 
dicular  AD :  the  cone  described  by 
jthe  triangle  ABD  is  the  third  part  of 
the  cylinder  described  by  the  rectan- 
gle AFBD  (Prop.  V.  Cor.) ;  also  the 
cone  described  by  the  triangle  ADC 
is  the  third  part  of  the  cylinder  de- 
sci-ibed  by  the  rectangle  ADCE  ;  hence  the  sum  of  the  two 
cones,  or  the  solid  described  by  ABC,  is  the  third  part  of  the 
two  cylinders  taken  together,  or  of  the  cylinder  described  by 
the  rectangle  BCEF. 


BOOK  VIII. 


179 


If  the  perpendicular  AD  falls  without 
the  triangle  ;  the  solid  described  by  ABC 
will,  in  that  case,  be  the  difference  of  the 
two  cones  described  by  ABD  and  ACD  ; 

but  at  the  same  time,  the  cylinder  de- . 

scribed  by  BCEF  will  be  the  difference  ^  ^     ^ 

of  the  two  cylinders  described  by  AFBD  and  AECD.  Hence 
the  solid,  described  by  the  revolution  of  the  triangle,  will  still 
be  a  third  part  of  the  cylinder  described  by  the  revolution  of 
the  rectangle  having  the  same  base  and  the  same  altitude. 

Scholium.  The  circle  of  which  AD  is  radius,  has  for  its 
measure  ^rx  AD-;  hence  t^x  AD^xBC  measures  the  cylinder 
described  by  BCEF,  and  ^ttx  AD-xBC  measures  the  solid 
described  by  the  triangle  ABC. 


PROPOSITION  XII.    LEMMA. 


^^tMJ^dtL^,:     . 


If  a  triangle  he  revolved  about  a  line  drawn  at  pleasure  through 
its  vertex,  the  solid  described  by  the  triangle  will  have  for  its 
measure,  the  area  of  the  triangle  multiplied  by  two  thirds  of  the 
circumference  traced,  by  the  middle  point  of  the  base. 

Let  CAB  be  the  triangle,  and  CD  the  line  about  which  it 
revolves. 

Produce  the  side  AB  till  it 
meets  the  axis  CD  in  D  ;  from  the 
points  A  and  B,  draw  AM,  BN, 
perpendicular  to  the  axis,  and  CP 
perpendicular  to  DA  produced. 

The  solid  described  by  the  tri- 
angle CAD  is  measured  by  \n  x 
AM^x  CD  (Prop.  XI.  Sch.) ;  the  solid  described  by  the  triangle 
CBD  is  measured  by  ^n  x  BN-  x  CD  ;  hence  the  difference  of 
those  solids,  or  the  solid  described  by  ABC,  will  have  for  its 
measure  i7r(AM2— BN^)  x  CD. 

To  this  expression  another  form  may  be  given.  From  I,  the 
middle  point  of  AB,draw  IK  perpendicular  to  CD  ;  and  through 
B,  draw  BO  parallel  to  CD :  we  shall  have  AM  +  BN=2lK 
(Book  IV.  Prop.  VII.) ;  andAM— BN=AO;  hence  (AM  + 
BN)  X*  ( AM— NB),  or  AM2_BN2=2IK  x  AO  (Book  IV.  Prop. 
X.).  Hence  the  measure  of  the  solid  in  question  is  ex- 
pressed by 

l^xIKxAOxCD. 


MX^sr 


18t/ 


OEOMETRf. 


M:x3>r 


But  CP  being  drawn  perpendicular  to  AB,  the  triangles  ABO 
DCP  will  be  similar,  and  give  the  proportion 

AO  :  CP  :  :  AB  :  CD; 
hence  AOxCD=CPxAB; 

but  CP  X  AB  is  double  the  area  of  the  triangle  ABC  ;  hence 
we  have 

A0xCD=2ABC; 

hence  the  solid  described  by  the 

triangle  ABC  is  also  measured 

by  |7r  X  ABC  X  IK,  or  which  is  the 

same  thing,  by  ABC  x  Icirc.  IK, 

arc.  IK  being  equal  to  S^rxIK. 

Hence  the  solid  described  by  the 

revolution  of  the  triangle  ABC, has 

for  its  measure  the  area  of  this  triangle  multiplied  hy  two  thirds 

of  the  circumference  traced  by  I,  the  middle  point  of  the  base. 

Cor,  IfthesideAC-=:CB, 
the  line  CI  will  be  perpen- 
dicular to  AB,  the  area  ABC 
will  be  equal  to  ABx|CI, 
and  the  solidity  ^n  x  ABC  x 
IK  will  become  fTixABx 
IKxCI.  But  the  triangles 
ABO,  CIK,  are  similar,  and 
give  the  proportion  AB  :  BO 
or  MN  :  :  CI  :  IK;  hence  ABxIK=MNxCI;  hence  the 
solid  described  by  the  isosceles  triangle  ABC  will  have  for  its 
measure  fTrxCFxMN  :  that  is,  equal  to  two  thirds  of  n  into 
the  square  of  the  perpendicular  let  fall  on  the  base,  into  the 
distance  between  the  pwo  perpendiculars  let  fall  on  the  axis. 


Scholium.  The  general  solution  appears  to  include  the  sup- 
position that  AB  produced  will  meet  the  axis  ;  but  the  results 
would  be  equally  true,  though  AB  were  parallel  to  the  axis. 

Thus,the  cylinder  described  by  AMNB   p  /\  j^ 

is  equal  to  tt.AM^.MN  ;  the  cone  descri- 
bed by  ACM  is  equal  to  In.KW.QM, 
and  the  cone  described  by  BCN  to 
^nkW  CN.  Add  the  first  two  solids  and 
take  away  the  third  ;  we  shall  have  the 
solid  described  by  ABC  equal  to  tt.AM^. 
(MN  +  iCM— iCN):  and  since  CN— CM =MN,  this  expres- 
sion  is  reducible  to  tt.AM-.^MN,  or  fTi.CP^.MN;  which  agreeg 
with  the  conclusion  found  above. 


BOOK  VIII. 


181 


PROPOSITION  XIII.    LEMMA. 


Ij  a  regular  semi-polygon  he  revolved  about  a  line  passing 
through  the  centre  and  the  vertices  of  iivo  opposite  angles,  the 
solid  described  will  be  equivalent  to  a  cone,  having  for  its  base 
the  inscribed  circle,  and  for  its  altitude  twice  the  axis  about 
which  the  semi-polygon  is  revolved. 


Let  the  semi-polygon  FABG  be  revolved 
about  FG  :  then,  if  01  be  the  radius  of  the 
inscribed  circle,  the  solid  described  will  be 
measured  by  ^area  01  x  2FG. 

For,  since  the  polygon  is  regular,  the 
triangles  OFA,  OAB,  OBC,  &c.  are  equal 
and  isosceles,  and  all  the  perpendiculars  let 
fall  from  O  on  the  bases  FA,  AB,  &c.  will 
be  equal  to  01,  the  radius  of  the  inscribed 
circle. 

Now,  the  solid  described  by  OAB  is  mea- 
sured by  frr  OP+MN  (Prop.  XII.  Cor.) ; 
the  solid  described  by  the  triangle  OFA  has  for  its  measure 
fTiOP  X  FM,  the  solid  described  by  the  triangle  OBC,  has  for 
its  measure  f  tiOP  x  NO,  and  since  the  same  may  be  shown  for 
the  solid  described  by  each  of  the  other  triangles,  it  follows 
that  the  entire  solid  described  by  the  semi-polygon  is  mea- 
sured by  |7iOP.(FM+MN  +  NO  +  OQ+QG),  or  f^OPxFG  ; 
which  is  also  equal  to  ^tiOP  x  2FG.  But  yr.OP  is  the  area  of 
the  inscribed  circle  (Book  V.  Prop.  XII.  Cor.  2.) :  hence  the 
solidity  is  equivalent  to  a  cone  whose  base  is  area  01,  and 
altitude  2FG. 


PROPOSITION  XIV.    THEOREM. 


The  solidity  of  a  sphere  is  equal  to  its  surface  multiplied  by  a 
third  of  its  radius. 


182  GEOMETRY. 

Inscribe  in  the  semicircle  ABCDE  a 
regular  semi-polygon,  having  any  number 
of  sides,  and  let  01  be  the  radius  of  the 
circle  inscribed  in  the  polygon. . 

If  the  semicircle  and  semi-polj^gon  be 
•'evolved  about  EA,  the  semicircle  will 
describe  a  sphere,  and  the  semi-polygon  a 
solid  which  has  for  its  measure  fyiOPx 
EA  (Prop.  XIII.) ;  and  this  will  be  true 
whatever  be  the  number  of  sides  of  the 

polygon.     But  if  the  number  of  sides  of  

the  polygon  be  indefinitely  increased,  the  E 

semi-polygon  will  become  the  semicircle,  01  will  become 
equal  to  OA,  and  the  solid  described  by  the  semi-polygon  will 
become  the  sphere  :  hence  the  solidity  of  the  sphere  is  equal 
to  fTiOA^xEA,  or  by  substituting  20 A  for  EA,  it  becomes 
jn.OA^  X  OA,  which  is  also  equal  to  47rOA2  x  ^OA.  But  47r.OA2 
is  equal  to  the  surface  of  the  sphere  (Prop.  X.  Cor.) :  hence 
the  solidity  of  a  sphere  is  equal  to  its  surface  multiphed  by  a 
third  of  its  radius. 

Scholium  1.  The  solidity  of  every  spherical  sector  is  equal  to 
the  zone  which  forms  its  base,  multiplied  by  a  third  of  the  radius. 

For,  the  solid  described  by  any  portion  of  the  regular  poly- 
gon, as  the  isosceles  triangle  OAB,  is  measured  by  f^rOFx  AF 
(Prop.  XII.  Cor.) ;  and  when  the  polygon  becomes  the  circle, 
the  portion-  OAB  becomes  the  sector  AOB,  01  becomes  equal 
to  OA,  and  the  solid  described  becomes  a  spherical  sector.  But 
its  measure  then  becomes  equal  to  f^r. AO^  x  AF,  which  is  equal 
to  27r.AO  X  AF  X  ^AO.  But  27r.AO  is  the  circumference  of  a 
great  circle  of  the  sphere  (Book  V.  Prop.  XII.  Cor.  2.),  which 
being  multiplied  by  AF  gives  the  surface  of  the  zone  which 
forms  the  base  of  the  sector  (Prop.  X.  Sch.  1.) :  and  the  proof 
is  equally  applicable  to  the  spherical  sector  described  by  the 
circular  sector  BOC  :  hence,  the  solidity  of  the  spherical  sector 
is  equal  to  the  zone  which  forms  its  base,  multiplied  by  a  third 
of  the  radius. 

Scholium  2.  Since  the  surface  of  a  sphere  whose  radius  is 
R,  is  expressed  by  47rR2  (Prop.  X.  Cor.),  it  follows  that  the 
surfaces  of  spheres  are  to  each  other  as  the  squares  of  their 
radii ;  and  since  their  solidities  are  as  their  surfaces  multiplied 
by  their  radii,  it  follows  that  the  solidities  of 'spheres  are  to 
each  other  as  the  cubes  of  their  radii,  or  as  the  cubes  of  their 
diameters. 


BOOK  VIII. 


183 


Scholium  3.  Let  R  be  the  radius  of  a  sphere  ;  its  surface 
will  be  expressed  by  47rR^,  and  its  solidity  by  AnK^  x  ^R,  or 
|7rR^.  If  the  diameter  is  called  D,  we  shall  have  R=iD, 
and  R'*'=|D^ :  hence  the  solidity  of  the  sphere  may  likewise  be 
expressed  by 


PROPOSITION  XV.    THEOREM. 


The  surface  of  a  sphere  is  to  the  whole  surface  of  the  circum- 
scribed cylinder^  including  its  bases,  as  2  is  to  3  :  and  the  so- 
lidities of  these  two  bodies  are  to  each  other  in  the  same  ratio. 


D 


^  «  ^ 

s^ 

^^^ 

/    "^ 

:    \ 

/^^— - 

^ 

0     ^ 

^ 

Let  MPNQ  be  a  great  circle  of  the 
sphere ;  ABCD  the  circumscribed 
square  :  if  the  semicircle  PMQ  and 
the  half  square  PADQ  are  at  the 
same  time  made  to  revolve  about  the 
diameter  PQ,  the  semicircle  will  gene-  ]M 
rate  the  sphere,  while  the  half  square 
will  generate  the  cylinder  circum- 
scribed about  that  sphere.  . 

The  altitude  AD  of  the  cylinder  is 
equal  to  the  diameter  PQ  ;  the  base  of 
the  cylinder  is  equal  to  the  great  circle,  since  its  diameter  AB 
is  equal  to  MN  ;  hence,  the  convex  surface  of  the  cylinder  is 
equal  to  the  circumference  of  the  great  circle  multiplied  by  its 
diameter  (Prop.  1.).  This  measure  is  the  same  as  that  of  the 
surface  of  the  sphere  (Prop.  X.) :  hence  the  surface  of  the  sphere 
is  efiual  to  the  convex  surface  of  the  circumscribed  cylinder. 

But  the  surface  of  the  sphere  is  equal  to  four  great  circles  ; 
hence  the  convex  surface  of  the  cylinder  is  also  equal  to  four 
great  circles  :  and  adding  the  two  bases,  each  equal  to  a  great 
circle,  the  total  surface  of  the  circumscribed  cylinder  will  be 
equal  to  six  great  circles;  hence  the  surface  of  the  sphere  is  to 
the  total  surface  of  the  circumscribed  cylinder  as  4  is  to  6,  or 
as  2  is  to  3  ;  which  was  the  first  branch  of  the  Proposition. 

In  the  next  place,  since  the  base  of  the  circumscribed  cylin- 
der is  equal  to  a  great  circle,  and  its  altitude  to  the  diameter, 
the  solidity  of  the  cylinder  will  be  equal  to  a  great  circle  mul- 
tiplied by  its  diameter  (Prop.  II.).  But  the  solidity  of  the 
sphere  is  equal  to  four  great  circles  multiplied  by  a  third  of  the 
radius  (Prop.  XIV.);  in  other  terms,  to  one  great  circle  multi- 
plied by  ^  of  the  radius,  or  by  |  of  the  diameter ;  hence  the 
sphere  is  to  the  circumscribed  cylinder  as  2  to  3,  and  conse- 
quently the  solidities  of  these  two  bodies  are  as  their  surface* 


184  GEOMETRY.  M 

% 
Scholium.  Conceive  a  polyedron,  all  of  whose  faces  touch 
the  sphere  ;  this  polyedron  may  be  considered  as  formed  of 
pyramids,  each  having  for  its  vertex  the  centre  of  the  sphere, 
and  for  its  base  one  of  the  polyedron's  faces.  Nov*^  it  is  evi- 
dent that  all  these  pyramids  will  have  the  radius  of  the  sphere 
for  their  common  altitude :  so  that  each  pyramid  will  be  equal 
to  one  face  of  the  polyedron  multiplied  by  a  third  of  the  radius ; 
hence  the  whole  polyedron  will  be  equal  to  its  surface  multi- 
plied by  a  third  of  the  radius  of  the  inscribed  sphere. 

It  is  therefore  manifest,  that  the  solidities  of  polyedrons  cir- 
cumscribed about  the  sphere  are  to  each  other  as  the  surfaces 
of  those  polyedrons.  Thus  the  property,  which  we  have  shown 
to  be  true  with  regard  to  the  circumscribed  cylinder,  is  also 
true  with  regard  to  an  infinite  number  of  other  bodies. 

We  might  likewise  have  observed  that  the  surfaces  of  poly- 
gons, circumscribed  about  the  circle,  are  to  each  other  as  their 
perimeters. 


PROPOSITION  XVI.    PROBLEM. 

If  a  circular  segment  he  supposed  to  make  a  revolution  about  a 
diameter  exterior  to  it^  required  the  value  of  the  solid  which  it 
describes,  A 


Let  the  segment  BMD  revolve  about  AC. 

On  the  axis,  let  fall  the  perpendiculars 
BE,  DF  ;  from  the  centre  C,  draw  CI 
perpendicular  to  the  chord  BD ;  also  draw 
the  radii  CB,  CD. 

The  solid  described  by  the  sector  BCD  .   C 

is  measured  by  f^  CB^.EF  (Prop.  XIV.  Sch.  1).  But  the 
solid  diescribed  by  the  isosceles  triangle  DCB  has  for  its  mea- 
sure fTT.CP.EF  (Prop.  XII.  Cor.) ;  hence  the  sohd  described 
by  the  segment  BMD=|^.EF.(CB2— CP).  Now,  in  the  right- 
angled  triangle  CBI,  we  have  CB^— CP^BP^iBD^  ;  hence 
the  solid  described  by  the  segment  BMD  will  have  for  its  mea- 
sure f  TT.EF.iBD^,  or  itt.BDIEF:  that  is  one  sixth  of  n  into 
the  square  of  the  chord,  into  the  distance  between  the  two  per- 
pendiculars let  fall  from  the  extremities  of  the  arc  on  the 
axis. 

Scholium.  The  solid  described  by  the  segment  BMD  is  to 
the  sphere  which  has  BD  for  its  diameter,  as  ^^r.BD-.EF  is 
to  i^.BD^  or  as  EF  to  BD. 


BOOK  VIII.  185 


PHOPOSITION  XVII.    THEOREM. 

Every  segment  of  a  sphere  is  measured  hy  the  half  sum  of 
its  bases  multiplied  hy  its  altitude,  plus  the  solidity  of  a 
sphere  whose  diameter  is  this  same  altitude. 


Let  BE,  DF,  be  the  radii  of  the  two 
bases  of  the  segment,  EF  its  altitude,  the 
segment  being  described  by  the  revolu- 
tion of  the  circular  space  BMDFE  about 
the  axis  FE.  The  solid  described  by  the 
segment  BMD  is  equal  to  ^rr.BD^.EF 
(Prop.  XVI.) ;  and  the  truncated  cone  de- 
scribed by  the  trapezoid  BDFE  is  equal 
to  i  7r.EF.  (BE2 + DF-+  BE.DF)  (Prop.V  I.) ; 
hence  the  segment  of  the  sphere,  which  is  the  sum  of  those  two 
solids,  must  be  equal  to  i7r.EF.(2BE2+2DF2+2BE.DF+BD:) 
But,  drawing  BO  parallel  to  EF,  we  shall  have  DO=DF— BE, 
hence  DO'^^DF^— 2DF.BE  +  BE2  (Book  IV.  Prop.  IX.)  ;  and 
consequently  BD2=zB02+ D0'^=:EF2+ DF^— 2DF .  BE  +  BE'^. 
Put  this  value  in  place  of  BD^  in  the  expression  for  the  value 
of  the  segment,  omitting  the  parts  which  destroy  each  other ; 
we  shall  obtain  for  the  solidity  of  the  segment, 

i7rEF.(3BE2+3DF2+EF2), 
an  expression  which  may  be  decomposed  into  two  parts  ;  the 

/7t.BE2+^.DF2\ 
one  i^.EF.(3BE2+3DF2),  or  EF.( ^ )  being  the 

rhalf  sum  of  the  bases  multiplied  by  the  altitude  ;  while  the 
other  iTt.EF^  represents  the  sphere  of  which  EF  is  the  diame- 
ter (Prop.  XIV.  Sch.) ;  hence  every  segment  of  a  sphere,  &c. 

;.  Cor,  If  either  of  the  bases  is  nothing,  the  segment  in  ques- 

%  tion  becomes  a  spherical  segment  with  a  single  base  ;  hence 

jl'  any  spherical  segment,  with  a  single  base,  is  equivalent  to  half 

§  Vie  cylinder  having  the  same  base  and  the  same  altitude,  plus  the 

'■k  sphere  of  which  this  altitude  is  the  diameter.. 

General  Scholium. 

Let  R  be  the  radius  of  a  cylinder's  base,  H  its  altitude  :  the 
I  .solidity  of  the  cylinder  will  be  ttR^  x  H,  or  ttR^H.  - 
^     Let  R  be  the  radius  of  a  cone's  base,  H  its  altitude:  the 
solidity  of  the  cone  will  be  nWx  ^H,  or  i^iR^H. 

Let  A  and  B  be  the  radii  of  the  bases  of  a  truncated  cone, 

a* 


186  GEOMETRY. 

H  its  altitude  :  the  solidity  of  the  truncated  cone  will  be  iTr.H. 
(A^+BHAB). 

Let  R  be  the  radius  of  a  sphere  ;  its  solidity  will  be  "flifR^. 

Let  R  be  the  radius  of  a  spherical  sector,  H  the  altitude  of 
the  zone,  which  forms  its  base  :  the  solidity  of  the  sector  will 
be  f  ^R2H. 

Let  P  and  Q  be  the  two  bases  of  a  spherical  segnient,  H  its 

altitude :  the  solidity  of  the  segment  will  be  -_I_z.H+|7r.H^. 

If  the  spherical  segment  has  but  one  base,  the  other  being 
nothing,  its  solidity  will  be  ^PH  +  ^tiH^. 

A 


BOOK  IX. 

OF  SPHERICAL  TRIANGLES  AND  SPHERICAL  POLYGONS. 


Definitions. 

1.  A  spherical  triangle  is  a  portion  of  the  surface  of  a  sphere, 
bounded  by  three  arcs  of  great  circles. 

These  arcs  are  named  the  sides  of  the  triangle,  and  are 
always  supposed  to  be  each  less  than  a  semi-circumference. 
The  angles,  which  their  planes  form  with  each  other,  are  the 
angles  of  the  triangle. 

2.  A  spherical  triangle  takes  the  name  of  right-angled, 
isosceles,  equilateral,  in  the  same  cases  as  a  rectilineal  triangle.^ 

3.  A  spherical  polygon  is  a  portion  of  the  surface  of  a  sphere  % 
terminfited  by  several  arcs  of  great  circles.  " 

4.  A  lune  is  that  portion  of  the  surface  of  a  sphere,  which  is 
included  between  two  great  sehii-circles  meeting  in  a  common 
diameter. 

5.  A  spherical  wedge  or  ungula  is  that  portion  of  the  solid 
sphere,  which  is  included  between  the  same  great  semi-circles, 
and  has  the  lune  for  its  base. 

6.  A  spherical  pyramid  is  a  portion  of  the  solid  sphere,  in- 
cluded between  the  planes  of  a  solid  angle  whose  vertex  is 
the  centre.  The  base  of  the  pyramid  is  the  spherical  polygon 
intercepted  by  the  same  planes. 

7.  The  fole  of  a  circle  of  a  sphere  is  a  point  in  the  surface 
equally  distant  from  all  the  points  in  the  circumference  of  this 
circle.  It  will  be  shown  (Prop.  V.)  that  every  circle,  great  or 
small,  has  always  two  poles. 


BOOK  IX.  187 

PROPOSITION  I.  THEOREM. 
In  every  spherical  triangle,  any  side  is  less  than  the  sttM  of  the 

other  two. 

Let  O  be  the  centre  of  the  sphere,  and 
ACB  the  triangle ;  draw  the  rad  i  i  O A,  OB, 
OC.  Imagine  the  planes  AOB,  AOC, 
COB,  to  be  drawn ;  these  planes  will  form 
a  solid  angle  at  the  centre  O ;  and  the  an- 
gles AOB,  AOC,  COB,  will  be  measured 
by  AB,  AC,  BC,  the  sides  of  the  spherical 
triangle.  But  each  of  the  three  plane  an- 
gles forming  a  solid  angle  is  less  than  the 
sum  of  the  other  two  (Book  VI.  Prop. 
XIX.) ;  hence  any  side  of  the  triangle 
ABC  is  less  than  the  sum  of  the  other  two. 

PROPOSITION  II.  THEOREM. 

The  shortest  path  from  one  point  to  another,  on  the  surface  of  a 
sphere,  is  the  arc  of  the  great  circle  which  joins  the  two  given 
points. 

Let  ANB  be  the  arc  of  a  great  circle 
which  joins  the  points  A  and  B ;  then  will  it 
be  the  shortest  path  between  them. 

1st.  If  two  points  N  and  B,  be  taken  on 
the  arc  of  a  great  circle,  at  unequal  distan- 
ces from  the  point  A,  the  shortest  distance 
from  B  to  A  will  be  greater  than  the  short- 
est distance  from  N  to  A.  ''b 

For,  about  A  as  a  pole  describe  a  circumference  CNP.  Now, 
the  line  of  shortest  distance  from  B  to  A  must  cross  this  circum- 
ference at  some  point  as  P.  But  the  shortest  distance  from  P  to 
A  whether  it  be  the  arc  of  a  great  circle  or  any  other  line,  is 
equal  to  the  shortest  distance  from  N  to  A;  for,  by  passing  the 
arc  of  a  great  circle  through  P  and  A,  and  revolving  it  about  the 
diameter  passing  through  A, the  point  P  maybe  made  to  coincide 
with  N,  when  the  shortest  distance  from  P  to  A  will  coincide 
with  the  shortest  distance  from  N  to  A  :  hence,  the  shortest  dis- 
tance from  B  to  A,  will  be  greater  than  the  shortest  distance 
from  N  to  A,  by  the  shortest  distance  from  B  to  P. 

If  the  point  B  be  taken  without  the  arc  AN,  still  making  AB 
greater  than  AN,  it  may  be  proved  in  a  mannerentirely  similar 
to  the  above,  that  the  shortest  distance  from  B  to  A  will  be  great- 
er than  the  shortest  distance  from  N  to  A. 

If  now,  there  be  a  shorter  path  between  the  points  B  and  A, 
than  the  arc  BDA  of  a  great  circle,  let  M  be  a  point  of  the  short- 


88  GEOMETRY. 

est  distance  possible;  then  through  M  draw  MA.  MB.  arcs  ot 
great  circles,  and  take  BD  equal  to  BM.  By  the  last  theorem, 
BDA<  BM  +  MA;  take  BD  =  BM  from  each,  and  there  will  re- 
main AD<  AM.  Now,  since  BM=:BD,  the  shortest  path  from  B 
to  M  is  equal  to  the  shortest  path  from  B  to  D:  hence  if  we  sup- 
pose two  paths  from  B  to  A,  one  passing  through  M  and  the  other 
through  D,  they  will  have  an  equal  part  in  each  ;  viz.  the  part 
from  B  to  M  equal  to  the  part  from  B  to  D. 
.  But  by  hypothesis,  the  path  through  M  is  the  shortest  path  from 
jB  to  A :  hence  the  shortest  path  from  M  to  A  must  be  less  than 
'ftre  shortest  path  from  D  to  A,  whereas  it  is  greater  since  the 
arc  MA  is  greater  than  DA :  hence,  no  point  of  the  shortest 
distance  between  B  and  A  can  lie  out  of  the  arc  of  the  great 
circle  BDA. 

I  *v- 

PROPOSITION  Illv  THEOREM. 

^The  sum  of  the  three  sides  of  a  spherical  triangle  is  less  than  the 
circumference  of  a  great  circle. 

Let  ABC  be  any  spherical  trian- 
gle ;  produce  the  sides  AB,  AC,  till 
they  meet  again  in  D.  The  arcs  ABD, 
ACD,  will  be  semicircumferences, 
since  two  great  circles  always  bisect 
each  other  (Book  VIII.  Prop.  VII. 
Cor.  2.).  But  in  the  triangle  BCD,  we 
have  the  side  BC<BD  +  CD  (Prop 
I.);  add  AB+AC  to  both;  we  shall 
have  AB  +  AC  +  BC<ABD  +  ACD, 
thatistosay,lessthanacircumference. 

PROPOSITION  IV.  THEOREM 

The  sum  of  all  the  sides  of  any  spherical  polygon  is  less  than  the 
circumference  of  a  great  circle. 

Take  the  pentagon  ABCDE,  for 
example.  Produce  the  sides  AB,  DC, 
till  they  meet  in  F;  then  since  BC  is 
less  than  BF  +  CF,  the  perimeter  of 
the  pentagon  ABCDE  will  be  less 
than  that  of  the  quadrilateral  AEDF. 

Again,  produce  the  sides  AE,  FD.  till  ^  A 

they  meet  in  G;  we  shall  have  ED<EG  +  DG;  hence  the  pe- 
rimeter of  the  quadrilateral  AEDF  is  less  than  that  of  the  tri- 
angle AFG  ;  which  last  is  itself  less  than  the  circumference  of 
a  great  circle  ;  hence,  for  a  still  stronger  reason,  the  perimeter 
of  the  polygon  ABCDE  is  less  than  this  same  circumference. 


L  C    Dt 


/^ 


BOOK  IX. 


189 


Scholium.  This  proposition  is  fun- 
damentally the  same  as  (Book  VI. 
Prop.  XX.) ;  for,  O  being  the  centre 
of  the  sphere,  a  sohd  angle  may  be 
conceived  as  formed  at  O  by  the  plane 
angles  AOB,  BOC,  COD,&c.,  and  the 
sum  of  these  angles  must  be  less  than 
four  right  angles  ;  which  is  exactly 
the  proposition  here  proved.     The  A 

demonstration  here  given  is  different  from  that  of  Book  VI. 
Prop.  XX. ;  both,  however,  suppose  that  the  polygon  ABCDE 
is  convex,  or  that  no  side  produced  will  cut  the  figure. 


PROPOSITION  V.    THEOREM. 

The  poles  of  a  great  circle  of  a  sphere,  are  the  extremities  of  that 
diameter  of  the  sphere  which  is  perpendicular  to  the  circle  ; 
and  these  extremities  are  also  the  poles  of  all  small  circles 
"parallel  to  it. 

Let  ED  be  perpendic- 
ular to  the  great  circle 
AMB  ;  then  will  E  and 
D  be  its  poles ;  as  also 
the  poles  of  the  parallel 
small  circles  HPI,FNG. 

For,  DC  being  per- 
pendicular to  the  plane 
AMB,  is  perpendicular 
to  all  the  straight  lines 
CA,  CM,  CB,&c.  drawn 
through  its  foot  in  this 
plane  ;  hence  all  the  arcs 
DA,  DM,  DB,  &c.  are 
quarters  of  the  circumfe- 
rence. So  likewise  are 
all  the  arcs  EA,  EM,  EB,  &c. ;  hence  the  points  D  and  E  are 
each  equally  distant  from  all  the  points  of  the  circumference 
AMB ;  hence,  they  are  the  poles  of  that  circumference  (Def.  7.). 

Again,  the  radius  DC,  perpendicular  to  the  plane  AMB,  is 
perpendicular  to  its  parallel  FNG ;  hence,  \\  passes  through  O 
the  centre  of  the  circle  FNG  (Book  VIII.  Prop.  VII.  Cor.  4.) ; 
hence,  if  the  oblique  lines  DF,  DN,  DG,  be  drawn,  these  ob- 
lique lines  will  diverge  equally  from  the  perpendicular  DO, 
and  will  themselves  be  equal.     But,  the  chords  being  equal, 


190 


GEOMETRY. 


ihe  arcs  are  equal ;  hence  the  point  D  is  the  pole  of  the  small 
circle  FNG  ;  and  for  like  reasons,  the  point  E  is  tlie  other  pole. 

Cor.  1.  Every  arc  DM, 
drawn  from  a  point  in 
the  arc  of  a  great  circle 
AMBto  its  pole,  is  a  quar- 
ter of  the  circumference, 
which  for  the  sake  of 
brevity,  is  usually  named 
a  quadrant  :  and  this 
quadrant  at  the  same 
time  makes  a  right  angle 
with  the  arc  AM.  For, 
the  line  DC  being  per- 
pendicular to  the  plane 
AMC,  every  plane  DME, 
passing  through  the  line 
DC  is  perpendicular  to 
the  plane  AMC  (Book  VI.  Prop.  XVI.);  hence,  the  angle  of 
these  planes,  or  the  angle  AMD,  is  a  right  angle. 

Cor.  2.  To  find  the  pole  of  a  given  arc  AM,  draw  the  indefi- 
nite arc  MD  perpendicular  to  AM  ;  take  MD  equal  to  a  quad- 
rant ;  the  point  D  will  be  one  of  the  poles  of  the  arc  AM :  or 
thus,  at  the  two  points  A  and  M,  draw  the  arcs  AD  and  MD 
perpendicular  to  AM ;  their  point  of  intersection  D  will  be  the 
pole  required. 

Cor.  3.  Conversely,  if  the  distance  of  the  point  D  from  each 
of  the  points  A  and  M  is  equal  to  a  quadrant,  the  point  D  will 
be  the  pole  of  the  arc  AM,  and  also  the  angles  DAM,  AMD, 
will  be  right  angles. 

For,  let  C  be  the  centre  of  the  sphere ;  and  draw  the  radii 
CA,  CD,  CM.  Since  the  angles  ACD,  MCD,  are  right  angles, 
the  line  CD  is  perpendicular  to  the  two  straight  lines  CA,  CM ; 
hence  it  is  perperpendicular  to  their  plane  (Book  VI.  Prop. 
IV,) ;  hence  the  point  D  is  the  pole  of  the  arc  AM  ;  and  conse- 
quently the  angles  DAM,  AMD,  are  right  angles. 

Scholium.  The  properties  of  these  poles  enable  us  to  describe 
arcs  ^of  a  circle  on  the  surface  of  a  sphere,  with  the  same 
facility  as  on  a  plane  surface.  It  is  evident,  for  instance,  that 
by  turning  the  arc  DF,  or  any  other  line  extending  to  the  same 
distance,  round  the  point  D,  the  extremity  F  will  describe  the 
small  circle  FNG ;  and  by  turning  the  quadrant  DFA  round 


BOOK  IX. 


191 


the  point  D,  its  extremity  A  will  describe  the  arc  of  the  great 
circle  AMB. 

If  the  arc  AM  were  required  to  be  produced,  and  nothing 
were  given  but  the  points  A  and  M  through  which  it  was  to 
pass,  we  should  first  have  to  determine  the  pole  D,  by  the 
intersection  of  two  arcs  described  from  the  points  A  and  M  as 
centres,  with  a  distance  equal  to  a  quadrant ;  the  pole  D  being 
found,  we  might  describe  the  arc  AM  and  its  prolongation, 
from  D  as  a  centre,  and  with  the  same  distance  as  before. 

In  fine,  if  it  be  required  from  a  given  point  P,  to  let  fall  a 
perpendicular  on  the  given  arc  AM ;  find,  a  point  on  the  arc 
AM  at  a  quadrant's  distance  from  the  point  P,  which  ig'done  by 
describing  an  arc  with  the  point  P  as  a  pole,  intersecting  AM  in  S : 
S  will  be  the  point  required,  and  is  the  pole  with  which  a  per- 
pendicular to  AM  may  be  described  passing  through  the  point  P. 


PROPOSITION  VI.    THEOREM. 

The  angle  formed  hy  two  arcs  of  great  circles^  is  equal  to  the  an- 
gle formed  by  the  tangents  of  these  arcs  at  their  point  of  inter- 
section, and  is  measured  by  the  arc  described  from  this  point 
of  intersection,  as  a  pole,  and  limited  by  the  sides,  produced  if 
necessary. 


O 


Let  the  angle  BAG  be  formed  by  the  two   A. 
arcs  AB,  AC  ;  then  will  it  be  equal  to  the 
angle  FAG  formed  by  the  tangents  AF,  AG, 
and  be  measured  by  the  arc  DE,  described 
about  A  as  a  pole. 

For  the  tangent  AF,  drawn  in  the  plane 
of  the  arc  AB,  is  perpendicular  to  the  radius 
AO ;  and  the  tangent  AG, drawn  in  the  plane 
of  the  arc  AC,  is  perpendicular  to  the  same 
radius  AO.  Hence  the  angle  FAG  is  equal 
to  the  angle  contained  by  the  planes  ABO, 
OAC  (Book  VI.  Def  4.) ;  which  is  that  of  Hi 
the  arcs  AB,  AC,  and  is  called  the  angle  BAG. 

In  like  manner,  if  the  arcs  AD  and  AE  are  both  quadrants, 
the  lines  OD,  OE,  will  be  perpendicular  to  OA,  and  the  angle 
DOE  will  still  be  equal  to  the  angle  of  the  planes  AOD,  AOE  : 
hence  the  arc  DE  is  the  measure  of  the  angle  contained  by 
these  planes,  or  of  the  angle  CAB. 

Cor.  The  angles  of  spherical  triangles  may  be  compared 
together,  by  means  of  the  arcs  of  great  circles  described  from 
their  vertices  as  poles  and  included  between  their  sides :  hence 
it  is  easy  to  make  an  angle  of  this  kind  equal  to  a  given  angle* 


192 


GEOMETRY. 


'Wr^' 


Scholium.  Vertical  angles,  such 
as  ACO  and  BCN  are  equal ;  for 
either  of  them  is  still  the  angle 
formed  by  the  two  planes  ACB, 
OCN. 

It  is  farther  evident,  that,  in  the 
intersection  of  two  arcs  ACB,  OCN, 
the  two  adjacent  angles  ACO,  OCB, 
taken  together,  are  equal  to  two 
right  angles. 


PROPOSITION  VII.     THEOREM. 


If  from  the  vertices  of  the  three  angles  of  a  spherical  triangle,  as 
poles,  three  arcs  he  described  forming  a  second  triangle,  the 
vertices  of  the  angles  of  this  second  triangle,  will  he  respectively 
poles  of  thei  sides' of  the  first. 

From  the  vertices  A,  B,  C, 
as  poles,  let  the  arcs  EF,  FD, 
ED,  be  described,  forming  on 
the  surface  of  the  sphere,  the 
triangle  DFE ;  then  will  the 
points  D,  E,  and  F,  be  respec- 
tively poles  of  the  sides  BC, 
AC,AB. 

For,  the  point  A  being  the 
pole  of  the  arc  EF,  the  dis- 
tance AE  is  a  quadrant ;  the 
point  C  being  the  pole  of  the  arc  DE,  the  distance  CE  is  like- 
wise a.  quadrant :  hence  the  point  E  is  removed  the  length  of  a 
quadrant  from  each  of  the  points  A  and  C  ;  hence,  it  is  the 
pole  of  the  arc  AC  (Prop.  V.  Cor.  3.).  It  might  be  shown,  by 
the  same  method,  that  D  is  the  pole  of  the  arc  BC,  and  F  that 
of  the  arc  AB. 

Cor,  Hence  the  triangle  ABC  may  be  descril)ed  by  means 
of  DEF,  as  DEF  is  described  by  means  of  ABC.  Triangles 
so  described  are  called  polar  triangles,  or  supplemental  tri- 
angles 


BOOK  IX.  193 

PROPOSITION  VIII.    THEOREM.  ^ 

The  same  supposition  continuing  as  in  the  last  Proposition^  each 
angle  in  one  of  the  triangles,  will  be  measured  by  a  semicir- 
cumference,  minus  the  side  lying  opposite  to  it  in  the  otiier 
triangle. 

For,  produce  the  sides  AB, 
AC,  if  necessary, till  they  meet 
EF,  in  G  and  H.  The  point  A 
being  the  pole  of  the  arc  GH, 
the  angle  A  will  be  measured 
by  that  arc  (Prop.  VI.).  But 
the  arc  EH  is  a  quadrant,  and 
likewise  GF,  E  being  the  pole 
of  AH,  and  F  of  AG  ;  hence 
EH  +  GF  is  equal  to  a  semi- 
circumference.  Now,  EH+  H 
GF  is  the  same  as  EF+GH  ;  hence  the  arc  GH,  which  mea- 
sures the  angle  A,  is  equal  to  a  semicircumference  minus  the 
side  EF.  In  like  manner,  the  angle  B  will  be  measured  by 
^^circ. — DF  :  the  angle  C,  by  |  circ. — DE. 

And  this  property  must  be  reciprocal  in  the  two  triangles, 
since  each  of  them  is  described  in  a  similar  manner  by  means 
of  the  other.  Thus  we  shall  find  the  angles  D,  E,  F,  of  the  triangle 
DEFtobe  measured  respectivelyby^  arc. — BC,  ^  circ. — AC, 
^  circ. — AB.  Thus  the  angle  D,  for  example,  is  measured  by 
the  arc  MI;  but  MI  +  BC=MC  +  BI=a  circ;  hence  the  are 
MI,  the  measure  of  D,  is  equal  to  ^  circ. — BC  ;  and  so  of  all 
the  rest. 

Scholium.  It  must  further  be  observed, 
that  besides  the  triangle  DEF,  three  others 
might  be  formed  by  the  intersection  of 
the  three  arcs  DE,  EF,  DF.  But  the 
proposition  immediately  before  us  is  ap- 
plicabJe  only  to  the  central  triangle, 
which  is  distinguished  from  the  other 
three  by  the  circumstance  (see  the  last 
figure)  that  the  two  angles  A  and  D  lie 
on  the  same  side  of  BC,  the  two  B  and  E  on  the  same  side  of 
AC,  and  the  two  C  and  F  on  the  same  side  of  AB. 

R 


104  GEOMETRY. 


y  PROPOSITION  IX.    THEOREM. 

£f  around  the  vertices  of  the  two  angles  of  a  given  spherical  tri- 
angle, as  poles,  the  circumferences  of  two  circles  be  described 
ivhich  shall  pass  through  the  third  angle  of  the  triangle;  if  then, 
through  the  other  point  in  which  these  circumferences  intersect 
and  the  two  first  angles  of  the  triangle,  the  arcs  of  great  cir- 
cles be  drawn,  the  triangle  thus  formed  will  have  all  its  parts 
equal  to  those  of  the  given  triangle. 

Let  ABC  be  the  given  triangle,  CED, 
DFC,  the  arcs  described  about  A  and  B 
as  poles ;  then  will  the  triangle  ADB  have 
all  its  parts  equal  to  those  of  ABC. 

For,  by  construction,  the  side  AD=r 
AC,  DB=BC,  and  AB  is  common  ;  hence 
these  two  triangles  have  their  sides  equal, 
each  to  each.  We  are  now  to  show,  that 
the  angles  opposite  these  equal  sides  are 
also  equal. 

If  the  centre  of  the  sphere  is  supposed  to  be  at  O,  a  solid 
angle  may  be  conceived  as  formed  at  O  by  the  three  plane 
angles  AOB,  AOC,  BOC  ;  likewise  another  solid  angle  may  be 
conceived  as  formed  by  the  three  plane  angles  AOB,  AOl), 
BOD.  And  because  the  sides  of  the  triangle  ABC  are  equal 
to  those  of  the  triangle  ADB,  the  plane  angles  forming  the  one 
of  these  solid  angles,  must  be  equal  to  the  plane  angles  forming 
the  other,  each  to  each.  But  in  that  case  we  have  shown  that 
the  planes,  in  which  the  equal  angles  lie,  are  equally  inclined 
to  each  other  (Book  VI.  Prop.  XXI.) ;  hence  all  the  angles  of 
the  spherical  triangle  DAB  are  respectively  equal  to  those  ot 
the  triangle  CAB,  namely,  DAB--BAC,  DBA=ABC,  and 
ADB  =  ACB;  hence  the  sides  and  the  angles  of  the  triangle 
ADB  are  equal  to  the  sides  and  the  angles  of  the  triangle  AC3. 

Scholium.  The  equality  of  these  triangles  is  not,  however, 
an  absolute  equality,  or  one  of  superposition  ;  for  it  would  be; 
impossible  to  apply  them  to  each  other  exactly,  unless  theyj 
were  isosceles.  The  equality  meant  here  is  what  we  have! 
already  named  an  equality  by  symmetry ;  therefore  w'e  shalJ 
call  the  triangles  ACB,  ADB,  symmetrical  tnangles. 


^»  BOOK  IX.  196 


PROPOSITION  X.    THEOREM. 

T\uo  triangles  on  the  same  sphere,  or  on  equal  spheres,  are  equal 
in  all  their  parts,  when  two  sides  and  the  included  angle  of  the 
one  are  equal  to  two  sides  and  the  included  angle  of  the  other^ 
each  to  each. 

Suppose  the  side  AB=EF,  the  side 
AC =EG,  and  the  angle  BACmFEG ; 
then  will  the  two  triangles  be  equal 
in  all  their  parts. 

For,  the  triangle  EFG  may  be 
placed  on  the  triangle  ABC,  or  on 
ABD  symmetrical  v/ith  ABC,  just  as 
two  rectilineal  triangles  are  placed 
upon  each  other,  when  they  have  an 
equal  angle  included  between  equal  sides.  Hence  all  the  parts 
of  the  triangle  EFG  will  be  equal  to  all  the  parts  of  the  trian- 
gle ABC  ;  that  is,  besides  the  three  parts  equal  by  hypothesis, 
we  shall  have  the  side  BC=FG,  the  angle  ABC = EFG,  and 
the  angle  ACB^EGF. 


PROPOSITION  XI.    THEOREM. 

Two  triangles  on  the  same  sphere,  or  on  equal  spheres,  are  equal 
in  all  their  parts,  when  two  angles  and  the  included  side  of  the 
one  are  equal  to  two  angles  and  the  included  side  of  the  other, 
each  to  each. 

For,  one  of  these  triangles,  or  the  triangle  symmetrical  with 
it,  may  be  placed  on  the  other,  as  is  done  in  the  corres- 
ponding case  of  rectilineal  triangles  (Book  I.  Prop.  VT.). 


PROPOSITION  XII.    THEOREM. 

If  two  triangles  on  the  same  sphere,  or  on  equal  spheres,  have  all 
their  sides  equal,  each  to  each,  their  angles  will  likewise  he 
equal,  each  to  each,  the  equal  angles  lying  opposite  the  equal 
sides. 


lOG 


GEOMETRY. 


This  truth  is  evident  from  Prop.  IX, 
where  it  was  shown,  that  with  three  given 
sides  AB,  AC,  BC,  there  can  only  be  two 
triangles  ACB,  ABD,  differing  as  to  the 
position  of  their  parts,  and  equal  as  to  the 
magnitude  of  those  parts.  Hence  those 
two  triangles,  having  all  their  sides  re- 
spectively equal  in  both,  must  either  be 
absolutely  equal,  or  at  least  symmetrically 
so  ;  in  either  of  which  cases,  their  corres- 
ponding angles  must  be  equal,  and  lie  opposite  to  equal  sides. 


PROPOSITION  XIII.    THEOREM. 


In  every  isosceles  spherical  triangle^  the  angles  opposite  the  equal 
sides  are  equal ;  and  conversely,  if  two  angles  of  a  spherical 
triangle  are  equal,  the  triangle  is  isosceles. 

First.  Suppose  the  side  AB  =  AC;  we  shall 
have  the  angle  C=B.  For,  if  the  arc  AD  be 
drawn  from  the  vertex  A  to  the  middle  point 
D  of  the  base,  the  two  triangles  ABD,  ACD, 
will  have  all  the  sides  of  the  one  respectively 
equal  to  the  corresponding  sides  of  the  other, 
namely,  AD  common,  BD=DC,  and  AB=: 
AC :  hence  by  the  last  Proposition,  their  an- 
gles will  be  equal ;  therefore,  B  =  C. 

Secondly.  Suppose  the  angle  B  =  C  ;  we  shall  have  the  side 
AC=AB.  For,  if  not,  let  AB  be  the  greater  of  the  two  ;  take 
BO=:AC,  and  draw  OC.  The  two  sides  BO,  BC,  are  equal  to 
the  two  AC,  BC ;  the  angle  OBC,  contained  by  the  first  two 
is  equal  to  ACB  contained  by  the  second  tv^^o.  Hence  the 
two  triangles  BOC,  ACB,  have  all  their  other  parts  equal 
(Prop.  X.) ;  hence  the  angle  OCB— ABC  :  but  by  hypothesis, 
the  angle  ABC=rACB ;  hence  we  have  OCB=ACB,  which  is 
absurd  ;  hence  it  is  absurd  to  suppose  AB  different  from  AC  ; 
hence  the  sides  AB,  AC,  opposite  to  the  equal  angles  B  and  C, 
are  equal. 

Scholium.  The  same  demonstration  proves  the  angle  BAD=^ 
DAC,  and  the  angle  BDA=ADC.  Hence  the  two  last  are 
right  angles  ;  hence  the  arc  drawn  from  the  vertex  of  an  isosceles 
spherical  triangle  to  the  middle  of  the  base,  is  at  right  angles  to 
that  base,  and  bisects  the  vertical  angle. 


BOOK  IX.  107 


rROlt)SITION  XIV.     THEOREM. 

In  any  spherical  triangle,  the  greater  side  is  opposite  the  greater 
angle ;  and  conversely,  the  greater  angle  is  opposite  the  greater 
side. 

Let  the  angle  A  be  greater 
than  the  angle  B,  then  will  BC 
be  greater  than  AC  ;  and  con- 
versely, if  BC  is  greater  than 
AC,  then  will  the  angle  A  be 
greater  than  B. 

First,  Suppose  the  angle  A>B  ;  make  the  angle  BAD=B ; 
then  we  shall  have  AD=DB  (Prop.  XIII.) :  but  AD  +  DC  is 
greater  than  AC  ;  hence,  putting  DB  in  place  of  AD,  we  shall 
haveDB  +  DCorBOAC. 

Secondly.  If  we  suppose  BC>AC,  the  angle  BAC  will  be 
greater  than  ABC.  For,  if  BAC  were  equal  to  ABC,  we 
should  have  BC=AC  ;  if  BAC  were  less  than  ABC,  we  should 
then,  as  has  just  been  shown,  find  BC<AC.  Both  these  con- 
clusions are  false :  hence  the  angle  BAC  is  greater  than  ABC. 

PROPOSITION  XV.    THEOREM. 

If  two  triangles  on  the  same  sphere,  or  on  equal  spheres,  are 
mutually  equiangular,  they  will  also  he  mutually  equilateral. 

Let  A  and  B  be  the  two  given  triangles  ;  P  and  Q  their  polar 
triangles.  Since  the  angles  are  equal  in  the  triangles  A  and 
B,  the  sides  will  be  equal  in.  their  polar  triangles  P  and  Q 
(Prop.  VIII.) :  but  since  the  triangles  P  and  Q  are  nnutually 
evuilateral,  they  must  also  be  mutually  equiangular  (Prop. 
XII.) ;  and  lastly,  the  angles  being  equal  in  the  triangles  P 
and  Q,  it  follows  that  the  sides  are  equal  in  their  polar  trian- 
gles A  and  B.  Hence  the  mutually  equiangular  triangles  A 
and  B  are  at  the  same  time  mutually  equilateral. 

Scholium.  This  proposition  is  not  applicable  to  rectilineal 
triangles ;  in  which  equality  among  the  angles  indicates  only 
proportionality  among  the  sides.  Nor  is  it  difficult  to  account 
for  the  difference  observable,  in  this  respect,  between  spherical 
and  rectilineal  triangles.    In  the  Proposition  now  before  us, 

R* 


J  98  GEOMETRY. 

as  well  as  in  the  preceding  ones,  which  treat  of  the  comparison 
of  triangles,  it  is  expressly  required  that  ihe  arcs  be  traced  on 
the  same  sphere,  or  on  equal  spheres.  Now  similar  arcs  are 
to  each  other  as  their  radii ;  hence,  on  equal  spheres,  two  tri- 
angles cannot  be  similar  without  being  equal.  Therefore  it  is 
not  strange  that  equality  among  the  angles  should  produce 
equality  among  the  sides. 

The  case  would  be  different,  if  the  triangles  were  drawn 
upon  unequal  spheres  ;  there,  the  angles  being  equal,  the  trian- 
gles would  be  similar,  and  the  homologous  sides  would  be  to 
each  other  as  the  radii  of  their  spheres. 


PROPOSITION  XVI.     THEOREM. 

The  sum  of  all  the  angles  in  any  spherical  triangle  is  less  than 
six  right  angles,  and  greater  than  two. 

For,  in  the  first  place,  every  angle  of  a  spherical  triangle  is 
less  than  two  right  angles  :  hence  the  sum  of  all  the  three  is 
less  than  six  right  angles. 

Secondly,  the  measure  of  each  angle  of  a  spherical  triangle 
is  equal  to  the  semicircumference  minus  the  corresponding  side 
of  the  polar  triangle  (Prop.  VIII.) ;  hence  the  sum  of  all  the  three, 
is  measured  by  the  three  semicircumferences  7ninusi\\e.  sum  of  all 
the  sides  of  the  polar  triangle.  Now  this  latter  sum  is  less  than  a 
circumference  (Prop.  III.) ;  therefore,  taking  it  away  from  three" 
semicircumferences,  the  remainder  will  be  greater  than  one 
semicircumference,  which  is  the  measure  of  two  right  angles ; 
hence,  in  the  second  place,  the  sum  of  all  the  angles  of  a  sphe- 
rical triangle  is  greater  than  two  right  angles. 

Cor.  1.  The  sum  of  all  the  angles  of  a  spherical  triangle  is 
not  constant,  like  that  of  all  the  angles  of  a  rectilineal  triangle  ; 
it  varies  between  two  right  angles  and  six,  without  ever  arriving 
at  either  of  these  limits.  Two  given  angles  therefore  do  not 
serve  to  determine  the  third. 

Cor.  2.  A  spherical  triangle  may  have  two,  or  even  three  of 
its  angles  right  angles ;  also  two,  or  even  threes  of  its  angles 
obtuse.  i^ 


BOOK  IX.  199 

Cor.  3.  If  the  triangle  ABC  is  hi-rectangular, 
in  ether  words,  has  two  right  angles  B  and  C, 
the  vertex  A  will  be  the  pole  of  the  base  BC  ; 
and  the  sides  AB,  AC,  will  be  quadrants 
(Prop.  V.  Cor.  3.). 

If  the  angle  A  is  also  a  right  angle,  the  tri-  ^ 
angle  ABC  will  be  iri-rectangular ;  its  angles  ^ 
will  all  be  right  angles,  and  its  sides  quadrants.  Two  of  the 
tri-rectangular  triangles  make  half  a  hemisphere,  four  make  a 
hemisphere,  and  the  tri-rectangular  triangle  is  obviously  con- 
tained eight  times  in  the  surface  of  a  sphere. 

Scholium.  In  all  the  preceding 
observations,  we  have  supposed,  in 
conformity  with  (Def.  1.)  that  sphe- 
rical triangles  have  always  each  of 
their  sides  less  than  a  semicircum- 
ference  ;  from  which  it  follows  that 
any  one  of  their  angles  is  always 
less  than  two  right  angles.  For,  if 
the  side  AB  is  less  than  a  semicir- 
cumference,  and  AC  is  so  likewise, 
both  those  arcs  will  require  to  be  E 

produced,  before  they  can  meet  in  D.  Now  the  two  angles 
ABC,  CBD,  taken  together,  are  equal  to  two  right  angles ; 
hence  the  angle  ABC  itself,  is  less  than  two  right  angles. 

We  may  observe,  however,  that  some  spherical  triangles  do 
exist,  in  which  certain  of  the  sides  are  greater  than  a  semicir- 
cumference,  and  certain  of  the  angles  greater  than  two  right 
angles.  Thus,  if  the  side  AC  is  produced  so  as  to  form  a  whole 
circumference  ACE,  the  part  which  remains,  after  subtracting 
the  triangle  ABC  from  the  hemisphere,  is  a  new  triangle  also 
designated  by  ABC,  and  having  AB,  BC,  AEDC  for  its  sides. 
Here,  it  is  plain,  the  side  AEDC  is  greater  than  the  semicir- 
cumferencc  AED  ;  and  at  the  same  time,  the  angle  B  opposite 
to  it  exceeds  two  right  angles,  by  the  quantity  CBD. 

The  trianjrles  whose  sides  and  angles  are  so  large,  have  been 
excluded  by  the  Definition  ;  but  the  only  reason  was,  that  the 
solution  of  them,  or  the  determmation  of  their  parts,  is  always 
reducible  to  the  solution  of  such  triangles  as  are  comprehended 
by  the  Definition.  Indeed,  it  is  evident  enough,  that  if  the  sides 
and  angles  of  the  triangle  ABC  are  known,  it  will  be  easy  to 
discover  the  angles  and  sides  of  the  triangle  which  bears  the 
same  name,  and  is  the  difference  between  a  hemisphere  and  the 
former  triangle. 


I 


200  GEOMETRY. 


PROPOSITION  XVII.    THEOREM. 

'llie  surface  of  a  lune  is  to  the  surface  of  the  sphere^  as  the  angle 
of  this  lune,  is  to  four  right  angles,  or  as  the  arc  which  mea- 
sures  that  angle,  is  to  the  circumference. 

Let  AMBN  be  a  lune  ;  then  will  its 
surface  be  to  the  surface  of  the  sphere 
as  the  angle  NCM  to  four  right  angles, 
or  as  the  arc  NM  to  the  circumference 
of  a  great  circle. 

Suppose,  in  the  first  place,  the  arc 
MN  to  be  to  the  circumference  MNPQ 
as  some  one  rational  number  is  to  ano- 
ther, as  5  to  48,  for  example.  The  cir- 
cumference MNPQ  being  divided  into 
48  equal  parts,  MN  will  contain  5  of  them  ;  and  if  the  pole  A 
were  joined  with  the  several  points  of  division,  by  as  many- 
quadrants,  we  should  in  the  hemisphere  AMNPQ  have  48  tri- 
angles, all  equal,  because  all  their  parts  are  equal.  Hence  the 
whole  sphere  must  contain  96  of  those  partial  triangles,  the  lune 
AMBNA  will  contain  10  of  them  ;  hence  the  lune  is  to  the 
sphere  as  10  is  to  96,  or  as  5  to  48,  in  other  words,  as  the  arc 
MN  is  to  the  circumference. 

If  the  arc  MN  is  not  commensurable  with  the  circumference, 
we  may  still  show,  by  a  mode  of  reasoning  frequently  exem- 
plified already,  that  in  that  case  also,  the  lune  is  to  the  sphere 
as  MN  is  to  the  circumference. 

Cor,  1.  Two  lunes  are  to  each  other  as  their  respective 
angles. 

Cor.  2.  It  was  shown  above,  that  the  whole  surface  of  the 
sphere  is  equal  to  eight  tri-rectangular  triangles  (Prop.  XVI. 
Cor.  3.) ;  hence,  if  the  area  of  one  such  triangle  is  represented 
by  T,  the  surface  of  the  whole  sphere  will  be  expressed  by  8T. 
This  granted,  if  the  right  angle  be  assumed  equal  to  l,the  sur- 
face of  the  lune  whose  angle  is  A,  will  be  expressed  by  2AxT: 
for, 

4:  A:  :  8T  :  2AxT 
in  which  expression,  A  represents  such  a  part  of  unity,  as  the 
angle  of  the  lune  is  of  one  right  angle. 

Scholium.  The  spherical  ungula,  bounded  by  the  planes 
AMB,  ANB,  IS  to  the  whole  solid  sphere,  as  the  angle  A  is  to 


BOOK  IX.  201 

four  right  angles.  For,  the  lunes  being  equal,  the  spherical 
ungulas  will  also  be  equal  ;  hence  two  spherical  ungulas  are  to 
each  other,  as  the  angles  formed  by  the  planes  which  bound 
them. 


PROPOSITION  XVIII.     THEOREM. 
Two  symmetrical  spherical  triangles  are  equivalent. 

Let  ABC,  DEF,  be  two  symmetri- 
cal triangles,  that  is  to  say,  two  tri- 
angles having  their  sides  AB=DE, 
AC=DF,  CB=EF,  and  yet  incapa- 
ble of  coinciding  with  each  other  :  /  I  \  q  p  / 
we  are  to  show  that  the  surface  ABC 
is  equal  to  the  surface  DEF. 

Let  P  be  the  pole  of  the  small 
circle  passing  through  the  three  points 
A,  B,  C  ;*  from  this  point  draw  the 
equal  arcs  PA,  PB,  PC  (Prop.  V.)  ;  at  the  point  F,  make  the 
angle  DFQzrACP,  the  arc  FQ=CP ;  and  draw  DQ,  EQ. 

The  sides  DF,  FQ,  are  equal  to  the  sides  AC,  CP  ;  the  an- 
gle DFQ=ACP :  hence  the  two  triangles  DFQ,  ACP  are  equal 
in  all  their  parts  (Prop.  X.) ;  hence  the  side  DQ=AP,  and  the 
angle  DQF=APC. 

In  the  proposed  triangles  DFE,  ABC,  the  angles  DFE,  ACB, 
opposite  to  the  equal  sides  DE,  AB,  being  equal  (Prop.  XII.). 
if  the  angles  DFQ?  ACP,  which  are  equal  by  construction,  be 
taken  ^way  from  them,  there  will  remain  the  angle  QFE,  equal 
to  PCB.  Also  the  sides  QF,  FE,  are  equal  to  the  sides  PC, 
CB ;  hence  the  two  triangles  FQE,  CPB,  are  equal  in  all  their 
parts  ;  hence  the  side  QE^PB,  and  the  angle  FQE  =  CPB. 

Now,  the  triangles  DFQ,  ACP,  which  have  their  sides  re- 
spectively equal,  are  at  the  same  time  isosceles,  and  capable  of 
coinciding,  when  applied  to  each  other;  for  having  placed  AC 
on  its  equal  DF,  the  equal  sides  will  fall  on  each  other,  and 
thus  the  two  triangles  will  exactly  coincide :  hence  they  are 
equal ;  and  the  surface  DQF— APC.  For  a  like  reason,  the 
surface  FQE=CPB,  and  the  surface  DQE=APB ;  hence  we 


»  The  circle  which  passes  through  the  three  points  A,  B,  C,  or  which  cir- 
cumscribes  the  triangle  ABC,  can  only  be  a  small  circle  of  the  sphere  ;  for  if 
it  were  a  great  circle,  the  three  sides  AB,  BC,  AC,  would  lie  in  one  plane,  and 
the  triangle  ABC  would  be  reduced  to  one  of  its  sides. 


202 


GEOMETRY. 


have    DQF+FQE— DQE=APC  +  CPB— APB,  or   DFE=: 

ABC ;  hence  the  two  symmetrical  triangles  ABC,  DEF  are 
equal  in  surface. 

Scholium,  The  poles  P  and  Q 
might  lie  within  triangles  ABC, 
DEF:  in  which  case  it  would  be 
requisite  to  add  the  three  triangles 


:)0  P.^ 


DQF,  FQE,  DQE,  together,  m  or- 
der to  make  up  the  triangle  DEF ; 
and  in  like  manner,  to  ?dd  the  three 
triangles  APC,  CPB,  APB,  together, 
in  order  to  make  up  the  triangle 
ABC  :  in  all  other  respects,  the  de- 
monstration and  the  result  would  still  be  the  same. 


PROPOSITION  XIX.    THEOREM. 


If  the  circumferences  of  two  great  circles  intersect  each  other  on 
the  surface  of  a  hemisphere,  the  sum  of  the  opposite  triangle^y 
thus  fo7^med,  is  equivalent  to  the  surface  of  a  lune  whose  angle 
is  equal  to  the  angle  formed  hy  the  circles. 

Let  the  circumferences  AOB,  COD, 
intersect  on  the  hemisphere  OACBD ; 
then  will  the  opposite  triangles  AOC, 
BOD,  be  equal  to  the  lune  whose  an- 
gle is  BOD. 

For,  producing  the  arcs  OB,  OD,  on 
the  other  hemisphere,  till  they  meet  in 
N,  the  arc  OBN  will  be  a  semi-circum- 
ference, and  AOB  one  also  ;  and  taking 
OB  from  each,  we  shall  have  BN= AO. 

For  a  like  reason,  we  have  DN=CO,  and  BD=AC.  He-nce, 
the  two  triangles  AOC,  BDN,  have  their  three  sides  respect- 
ively equal ;  they  are  therefore  symmetrical ;  hence  they  are 
equal  in  surface  (Prop.  XVIII.) :  but  the  sum  of  the  triangles 
BDN,  BOD,  is  equivalent  to  the  lune  OBNDO, 'whose  angle  is 
BOD:  hence,  AOC  +  BOD  is  equivalent  to  the  lune  whose 
angle  is  BOD. 

Scholium.  It  is  likewise  evident  that  the  two  spherical  pyra- 
mids, which  have  the  triangles  AOC,  BOD,  for  bases,  are  toge- 
ther equivalent  to  the  spherical  ungula  whose  angle  is  BOD. 


BOOK  IX.  203 

PROPOSITION  XX.  THEOREM. 

The  surface  of  a  spherical  triangle  is  measured  by  the  excess  of 
the  sum  of  its  three  angles  above  two  right  angles,  multiplied 
by  the  tri-rectangular  triangle. 

Let  ABC  be  the  proposed  triangle  :  pro- 
duce its  sides  till  they  meet  the  great  circle 
DEFG  drawn  at  pleasure  without  the  trian- 
gle. By  the  last  Theorem,  the  two  triangles 
ADE,  AGH,  are  together  equivalent  to  the 
lune  whose  angle  is  A,  and  which  is  mea- 
sured by  2A.T  (Prop.  XVII.  Cor.  2.). 
Hence  we  have  ADE  + AGH=2A.T  ;  and 
for  a  like  reason,  BGF+BID  =  2B.T,  and 
CIH  +  CFE=2C.T  But  the  sum  of  these 
six  triangles  exceeds  the  hemisphere  by  twice  the  triangle 
ABC,  and  the  hemisphere  is  represented  by  4T ;  therefore, 
twice  the  triangle  ABC  is  equal  to  2A.T  +  2B.T  +  2C.T— 4  T; 
and  consequently,  once  ABC  =  (A  +  B-j-C — 2)T;  hence  every 
spherical  triangle  is  measured  by  the  sum  of  all  its  angles  minus 
two  right  angles,  multiplied  by  the  tri-rectangular  triangle. 

Coj\  1.  However  many  right  angles  there  may  be  in  the  sum  of 
the  three  angles  minus  two  right  angles,just  so  many  tri-rectan- 
gular triangles,  or  eighths  of  the  sphere,  will  the  proposed  trian- 
gle contain.  If  the  angles,  for  example,  are  each  equal  to  f  of 
a  right  angle,  the  three  angles  will  amount  to  4  right  angles,  and 
the  sum  of  the  angles  minus  two  right  angles  will  be  represented 
by  4 — 2  or  2;  therefore  the  surface  of  the  triangle  will  be  equal 
to  two  tri-rectangular  triangles,  or  to  the  fourth  part  of  the 
whole  surface  of  the  sphere. 

Scholium,  While  the  spherical  triangle  ABC  is  compared 
with  the  tri-rectangular  triangle,  the  spherical  pyramid,  which 
has  ABC  for  its  base,  is  compared  with  the  tri-rectangular  py- 
ramid, and  a  similar  proportion  is  found  to  subsist  between 
them.  The  solid  angle  at  the  vertex  of  the  pyramid,  is  in  like 
manner  compared  with  the  solid  angle  at  the  vertex  of  the  tri- 
rectangular  pyramid.  These  comparisons  are  founded  on  the 
coincidence  of  the  corresponding  parts.     If  the  bases  of  tho 


JiOl  GEOMETRY. 

pyramids  coincide,  the  pyramids  themselves  will  evidently  co- 
incide, and  likewise  the  solid  angles  at  their  vertices.  From 
tliis,  some  consequences  are  deduced. 

First.  Two  triangular  spherical  pyramids  are  to  each  other 
as  their  bases  :  and  since  a  polygonal  pyramid  may  always  be 
divided  into  a  certain  number  of  triangular  ones,  it  follows  that 
any  two  spherical  pyramids  are  to  each  other,  as  the  polygons 
which  form  their  bases. 

Second.  The  solid  angles  at  the  vertices  of  these  pyramids,  are 
also  as  their  bases  ;  hence,  for  comparing  any  two  solid  angles, 
we  have  merely  to  place  their  vertices  at  the  centres  of  two 
equal  spheres,  and  the  solid  angles  will  be  to  each  other  as  the 
spherical  polygons  intercepted  between  their  planes  or  faces. 

The^  vertical  angle  of  the  tri-rectangular  pyramid  is  formed 
by  three  planes  at  right  angles  to  each  other  :  this  angle,  which 
may  be  called  a  right  solid  aiigle,  will  serve  as  a  very  natural 
unit  of  measure  for  all  other  solid  angles.  If,  for  example,  the 
the  area  of  the  triangle  is  f  of  the  tri-rectangular  triangle, 
then  the  corresponding  solid  angle  will  also  be  f  of  the 
right  solid  an^le. 

PROPOSITION  XXI.  THEOREM 

The  surface  of  a  spherical  polygon  is  measured  by  the  sum  of  all 
its  angles,m\n\is  two  right  angles  multiplied  by  the  number  of 
sides  in  the  polygon  less  two,  into  the  tri-rectangular  triangle. 

From  one  of  the  vertices  A,  let  diago- 
nals AC,  AD  be  drawn  to  all  the  other  ver- 
tices ;  the  polygon  ABCDE  will  be  di- 
vided into  as  many  triangles  minus  two  as 
it  has, sides.  But  the  surface  of  each  tri- 
angle is  measured  by  the  sum  of  all  its  an- 
gles minus  two  right  angles,  into  the  tri- 
rectangular  triangle ;  and  the  sum  of  the  angles  in  all  the  tri- 
angles is  evidently  the  same  as  that  of  all  the  angles  of  the 
polygon ;  hence,  the  surface  of  the  polygon  is  equal  to  the  sum 
of  all  its  angles, diminished  by  twice  as  many  pght  angles  as 
it  has  sides   less  two,    into  the  tri-rectangular  triangle. 

Scholium.  Let  s  be  the  sum  of  all  the  angles  in  a  spherical 
polygon,  n  the  number  of  its  sides,  and  T  the  tri-rectangular  tri- 
angle ;  the  right  angle  being  taken  for  unity,  the  surface  of  the 
polygon  will  be  measured  by 

(s-^2  (n— 2,))  T,  or  (s— 2  w^4)  T 


APPENDIX, 


THE  REGULAR  POLYEDRONS. 


A  regular  polyedron  is  one  whose  faces  are  all  equal  regular 
polygons,  and  whose  solid  angles  are  all  equal  to  each  other. 
There  are  five  such  polyedrons. 

First.  If  the  faces  are  equilateral  triangles,  polyedrons  may 
be  formed  of  them,  having  solid  angles  contained  by  three  of 
those  triangles,  by  four,  or  by  five  :  hence  arise  three  regular 
bodies,  the  tetraedron,  the  octaedron,  the  icosaedron.  No  other 
can  be  formed  with  equilateral  triangles  ;  for  six  angles  of  such 
a  triangle  are  equal  to  four  right  angles,  and  cannot  form  a 
solid  angle  (Book  VI.  Prop.  XX.). 

Secondly.  If  the  faces  are  squares,  their  angles  may  be  ar- 
ranged by  threes :  hence  results  the  hexaedron  or  cube.  Four 
angles  of  a  square  are  equal  to  four  right  angles,  and  cannot 
form  a  solid  angle. 

Thirdly.  In  fine,  if  the  faces  are  regular  pentagons,  their 
angles  likewise  may  be  arranged  by  threes :  the  regular  dode- 
caedron  will  result. 

We  can  proceed  no  farther  :  three  angles  of  a  regular  hexa- 
gon are  equal  to  four  right  angles ;  three  of  a  heptagon  are 
greater. 

Hence  there  can  only  be  five  regular  polyedrons ;  three  formed 
with  equilateral  triangles,  one  with  squares,  and  one  with  pen- 
tagons. 

Construction  of  the  Tetraedron, 

Let  ABC  be  the  equilateral  triangle 
which  is  to  form  one  face  of  the  tetrae- 
dron. At  the  point  O,  the  centre  of  this 
triangle,  erect  OS  perpendicular  to  the 
plane  ABC  ;  terminate  this  perpendicular 
in  S,  so  that  AS=AB;  draw  SB,  SC : 
the  pyramid  S-ABC  will  be  the  tetrae- 
dron required. 

For,  by  reason  of  the  equal  distances 
OA,  OB,  OC,  the  oblique  lines  SA,  SB,  SC,  are  equally  re- 

S 


206 


APPENDIX. 


moved  from  the  perpendicular  SO,  and 
consequently  equal  (Book  VI.  Prop.  V.). 
One  of  them  SA=AB ;  hence  the  four 
faces  of  the  pyramid  S-ABC,  are  trian- 
gles, equal  to  the  given  triangle  ABC. 
And  the  solid  angles  of  this  pyramid 
are  all  equal,  because  each  of  them  is 
formed  by  three  equal  plane  angles: 
hence  this  pyramid  is  a  regular  tetrae- 
dron. 


Construction  of  the  Hexaedron. 

Let  ABCD  be  a  given  square.  On  the 
base  ABCD,  construct  a  right  prism  whose 
altitude  AE  shall  be  equal  to  the  side  AB. 
The  faces  of  this  prism  will  evidently  be 
equal  squares ;  and  its  solid  angles  all  equal, 
each  being  formed  with  three  right  angles  :  ■ 
hence  this  prism  is  a  regular  hexaedron  or 
cube. 


E 


C\ 


T? 


13 


The  following  propositions  can  be  easily  proved. 

1."  Any  regular  polyedron  may  be  divided  into  as  many 
regular  pyramids  as  the  polyedron  has  faces ;  the  common 
vertex  of  these  pyramids  will  be  the  centre  of  the  polyedron  ; 
and  at  the  same  time,  that  of  the  inscribed  and  of  the  circum- 
scribed sphere. 

2.  The  solidity  of  a  regular  polyedron  is  equal  to  its  sur- 
face multiplied  by  a  third  part  of  the  radius  of  the  inscribed 
sphere. 

3.  Two  regular  polyedrons  of  the  same  name,  are  two  simi- 
lar sojids,  and  their  homologous  dimensions  are  proportional ; 
hence  the  radii  of  the  inscribed  or  the  circumscribed  spheres 
are  to  each  other  as  the  sides  of  the  polyedrons. 

4.  If  a  regular  polyedron  is  inscribed  in  a  sphere,  the  planes 
drawn  from  the  centre,  through  the  different  edges,  will  divide 
the  surface  of  the  sphere  into  as  many  spherical  polygons,  all 
equal  and  similar,  as  the  polyedron  has  faces. 


APPLICATION  OF  ALGEBRA. 


TO  THE  SOLUTION  OF 


GEOMETRICAL  PROBLEMS. 


A  problem  is  a  question  which  requires  a  solution.  A  geo- 
metrical problem  is  one,  in  which  certain  parts  of  a  geometri- 
cal figure  are  given  or  known,  from  which  it  is  required  to  de- 
termine certain  other  parts. 

When  it  is  proposed  to  solve  a  geometrical  problem  by 
means  of  Algebra,  the  given  parts  are  represented  by  the  first 
letters  of  the  alphabet,  and  the  required  parts  by  the  final  let- 
ters, and  the  relations  which  subsist  between  the  known  and 
unknown  parts  furnish  the  equations  of  the  problem.  The  solu- 
tion of  these  equations,  when  so  formed,  gives  the  solution  of 
the  problem. 

No  general  rule  can  be  given  for  forming  the  equations.  The 
equations  must  be  independent  of  each  other,  and  their  number 
equal  to  that  of  the  unknown  quantities  introduced  (Alg. 
Art.  103.).  Experience,  and  a  careful  examination  of  all  the 
conditions,  whether  explicit  or  implicit  (Alg.  Art.  94,)  will 
serve  as  guides  in  stating  the  questions ;  to  which  may  be 
added  the  following  particular  directions. 

1st.  Draw  a  figure  which  shall  represent  all  the  given  parts, 
and  all  the  required  parts.  Then  draw  such  other  lines  as  will 
establish  the  most  simple  relations  between  them.  If  an  angle 
is  given,  it  is  generally  best  to  let  fall  a  perpendicular  that  shall 
lie  opposite  to  it;  and  this  perpendicular,  if  possible,  should  be 
drawn  from  the  extremity  of  a  given  side. 

2d.  When  two  lines  or  quantities  are  connected  in  the  same 
way  with  other  parts  of  the  figure  or  problem,  it  is  in  general, 
not  best  to  use  either  of  them  separately;  but  to  use  their  sum, 
their  difference,  their  product,  their  quotient,  or  perhaps  ano- 
ther line  of  the  figure  with  which  they  are  alike  connected. 

3d.  When  the  area,  or  perimeter  of  a  figure,  is  given,  it  is 
sometimes  best  to  assume  another  figure  similar  to  the  propo- 
sed, having  one  of  its  sides  equal  to  unity,  or  some  other  known 
quantity.  A  comparison  of  the  two  figures  will  often  give  a  re- 
quired part.     We  will  add  the  following  problems.* 

*  The  following  problems  are  selected  from  Hutton's  Application  of  Algebra 
to  Geometry,  and  the  examples  in  Mensuration  from  his  treatise  on  that  subject. 


208  APPLICATION  OF  ALGEBRA 


PROBLEM   I. 

In  a  right  angled  triangle  BAG,  having  given  the  base  BA, 
and  the  sum  of  the  hypothenuse  and  perpendicular,  it  is  re- 
quired to  find  the  hypothenuse  and  perpendicular. 

Put   BA=c=3,  BC=a:;,  KC=y   and  the  sum  of  the  hypo- 
thenuse and  perpendicular  equal  to     s  =  9 

Then,  x-^y=s=Q. 

and       x'=y''+c^    (Bk  .  IV.  Prop.  XL) 
From  1st  equ:   x=s — y 

and  x^= s^ — 2sy  -\-y^  p 

By  subtracting,  0  =  s^ — 2sy — c^ 

or  2sy=s'^ — c^ 

s'^—c^ 
hence,  y=-^  =4=Ae 

Therefore  a; +  4 =9    or    a:=5=BC. 

PROBLEM  II. 

In  a  Hght  angled  triangle,  having  given  the  hypothenuse,  and  ike 
sum  of  the  base  and  perpendicular,  to  find  these  two  sides' 

Put  BC=a=5,  BA=a:,  AC=y         and  the  sum 
of  the  base  and  perpendicular =5=7 
Then  x-\-y=s=l 

and  a;2+3/^=ar^ 

From  first  equation        x=s — y 

or  x^=s'^—2sy-{-y'^ 

Hence,  y^==a^-.s'2+2sy^y^ 

or  2?/^ — 2sy=a^—s^ 

or  'ir—sy-. 


2 

By  completing  the  square  y^ — 5?/  +  i5^=|a^ — \s^ 

or  y =is±  yia^— {5-=4  or  3 

Hence  a:= J5=F  Vid-—\s'^='^  or  4 


TO  GEOMETRY. 


209 


PROBLEM  III. 

In  a  rectangle^  having  given  the  diagonal  and  perimeter,  to  find 

the  sidea. 


Let  ABCD  be   the  proposed  rectangle. 
Put  AC=^=10,  the  perimeter = 2a  :^ 28,    or 
AB  +  BC=a=14:  also  put  AB=a;and  BC=y. 

Then,  x^+y'=d^ 

and  x+y=a 

From  which  equations  we  obtain, 


=ia=fc  Vld^—ia^=8  or  6, 


and 


a;=iflr=p  V^d^—ia^=6  or  8. 


PROBLEM  IV. 

Having  given  the  base  and  perpendicular  of  a  triangle^  to  find 
the  side  of  an  inscribed  square. 

Let  ABC  be  the  triangle  and  HEFG 
the  inscribed  square.  Put  AB=b,  CD=a, 
and  HE  or  GHrra; :    then  Cl=a — x. 
We  have  by  similar  triangles 

AB:  CD:;  GF:  CI 
or  b:  a::  x:  a — x 

Hence,         ab — bx=ax 
ab 


or 


a  +  b 


the  side   of   the  inscribed  square ; 


which,  therefore,  depends  only  on  the  base  and  altitude  of  the 
triangle. 


PROBLEM  V. 

In  an  equilateral  triangle,  having  given  the  lengths  of  the 
three  perpendiculars  drawn  from  a  point  within,  on  the  three 
sides:  to  determine  the  sides  of  the  triangle. 


210  APPLICATION  OF  ALGEBRA 

Let  ABC  be  the  equilateral  triangle ; 
DG,  DE  and  DF  the  given  perpendicu- 
lars hi  fall  from  D  on  the  sides.  Draw 
DA,  DB,  DC  to  the  vertices  of  the  angles, 
and  let  fall  the  perpendicular  CH  on 
the  base.  Let  DG=a,  J)E=b,  and 
DF=c :  put  one  of  the  equal  sides  AB 

=2x;    hence   AH=x,    and   CH= 

Now  since  the  area  of  a  triangle  is  equal  to  half  its  base 
into  the  altitude,  (Bk.  IV.  Prop.  VL) 

iAB  X  CH=a;  x  x  Vs^x^  -/s^triangle  ACB 

iAB  X  DG=.r  X  a      =ax      =  triangle  ADB 

iBCxJ)E=xxb       =bx      =triangle  BCD 

^ACxDF=a;xc       =cx     =triangle  ACD 

But  the  three  last  triangles  make  up,  and  are  consequently 
equal  to,  the  first ;  hence, 

x^  V3=ax  +  bx  +  cx^=x{a  +  6  +  c) ; 

or  xV3=a-\-b  +  c 

a-\-b  +  c 


therefore", 


V3 


Remark.     Since  the  perpendicular  CH  is  equal  to  xy  3,  it  ^ 

is  consequently  equal  to  «  +  6  +  c;  that  is,  the  perpendicular  let  I 

fall  from  either  angle  of  an  equilateral  triangle  on  the  oppo-  • 

site  side,  is  equal  to  the  sum  of  the  three  perpendiculars  let  " 

fall  from  any  point  within  the  triangle  on  the  sides  respectively.  ^ 


PROBLEM   VI. 


In  a  right  angled  triangle,  having  given  the  base  and  the  dif- 
ference between  the  hypothenuse  and  perpendicular,  to  find 
the  sides. 


PROBLEM   VII. 


In  a  right  angled  triangle,  having  given  the  hypothenuse  and 
the  difference  between  the  base  and  perpendicular,  to  deter- 


mine the  triangle. 


► 


TO  GEOMETRY.  211 

PROBLEM    VIII. 

Having  given  the  area  of  a  rectangle  inscribed  in  a  given 
triangle  ;  to  determine  the  sides  of  the  rectangle. 

PROBLEM    IX. 

In  a  triangle,  having  given  the  ratio  of  the  two  sides,  togeth- 
er with  both  the  segments  of  the  base  made  by  a  perpendic- 
ular from  the  vertical  angle  ;  to  determine  the  triangle. 

PROBLEM    X. 

In  a  triangle,  having  given  the  base,  the  sum  of  the  other  two 
sides,  and  the  length  of  a  line  drawn  from  the  vertical  angle 
to  the  middle  of  the  base  ;  to  find  the  sides  of  the  triangle. 

PROBLEM   XI. 

In  a  triangle,  having  given  the  two  sid6s  about  the  vertical 
angle,  together  with  the  line  bisecting  that  angle  and  terminating 
in  the  base  ;  to  find  the  base. 

PROBLEM   XII. 

To  determine  a  right  angled  triangle,  having  given  the 
lengths  of  two  lines  drawn  trom  the  acute  angles  to  the  mid^ 
die  of  the  opposite  sides. 

PROBLEM   Xni. 


I 


To  determine  a  right-angled  triangle,  having  given  the  pe- 
rimeter and  the  radius  of  the  inscribed  circle. 


PROBLEM   XIV. 


To  determine  a  triangle,  having  given  the  base,  the  per- 
pendicular and  the  ratio  of  the  two  sides. 

PROBLEM    XV. 

To  determine  a  right  angled  triangle,  having  given  the 
hypothenuse,  and  the  side  of  the  inscribed  square. 

PROBLEM    XVI. 

To  determine  the  radii  of  three  equal  circles,  described 
within  and  tangent  to,  a  given  circle,  and  also  tangent  to 
each  other. 


212  APPLICATION  OF  ALGEBRA 


PROBLEM   XVII 


In  a  right  angled  triangle,  having  given  the  perimeter  and 
the  perpendicular  let  fall  from  the  right  angle  on  the  hypothe- 


nuse,  to  determine  the  triangle. 


PROBLEM   XVIII. 


To  determine  a  right  angled  triangle,  having  given  the 
hypothenuse  and  the  difference  of  two  lines  drawn  from  the 
two  acute  angles  to  the  centre  of  the  inscribed  circle. 


PROBLEM    XIX. 


To  determine  a  triangle,  having  given  the  base,  the  perpen- 
dicular, and  the  difference  of  the  two  other  sides. 


PROBLEM    XX. 


To  determine  a  triangle,  having  given  the  base,  the  perpen- 
dicular and  the  rectangle  of  the  two  sides. 


PROBLEM    XXI. 


To  determine  a  triangle,  having  given  the  lengths  of  three 
lines  drawn  from  the  three  angles  to  the  middle  of  the  opposite 
sides. 


PROBLEM    XXII. 


In  a  triangle,  having  given  the  three  sides,  to  find  the  radius 
of  the  inscribed  circle. 


PROBLEM   XXIII. 


To  determine  a  right  angled  triangle,  having  given  the  side 
of  the  inscribed  square,  and  the  radius  of  the  inscribed  circle. 


PROBLEM   XXIV. 


To  determine   a  right  angled   triangle,  having  given  the 
hypothenuse  and  radius  of  the  inscribed  circle. 


PROBLEM   XXV. 


To  determine  a  triangle,  having  given  the  base,  the  line 
bisecting  the  vertical  angle,  and  the  diameter  of  the  circum- 
scribing circle. 


PLANE  TRIGONOMETRY.        213 


PLANE  TRIGONOMETRY. 

In  every  triangle  there  are  six  parts :  three  sides  and  three 
angles.  These  parts  are  so  related  to  each  other,  that  if  a 
certain  number  of  them  be  known  or  given,  the  remaining 
ones  can  be  determined. 

Plane  Trigonometry  explains  the  methods  of  finding,  by  cal- 
culation, the  unknown  parts  of  a  rectilineal  triangle,  when 
a  sufficient  number  of  the  six  parts  are  given. 

When  three  of  the  six  parts  are  known,  and  one  of  them  is  a 
side,  the  remaining  parts  can  always  be  found.  If  the  three 
angles  were  given,  it  is  obvious  that  the  problem  would  be  in- 
determinate, since  all  similar  triangles  would  satisfy  the  con- 
ditions. 

It  has  already  been  shown,  in  the  problems  annexed  to  Book 
III.,  how  rectilineal  triangles  are  constructed  by  means  of  three 
given  parts.  But  these  constructions,  which  are  called  graphic 
methods,  though  perfectly  correct  in  theory,  would  give  only 
a  moderate  approximation  in  practice,  on  account  of  the  im- 
perfection of  the  instruments  required  in  constructing  them. 
Trigonometrical  methods,  on  the  contrary,  being  independent 
of  all  mechanical  operations,  give  solutions  with  the  utmost 
accuracy. 

These  methods  are  founded  upon  the  properties  of  lines  called 
trigonometrical  lines,  which  furnish  a  very  simple  mode  of  ex- 
pressing the  relations  between  the  sides  and  angles  of  triangles. 

We  shall  first  explain  the  properties  of  those  lines,  and  the 
principal  forraulas  derived  from  them  ;  formulas  which  are  of 
great  use  in  all  the  branches  of  mathematics,  and  which  even 
furnish  means  of  improvement  to  algebraical  analysis.  We 
shall  next  apply  those  results  to  the  solution  of  rectilineal  tri- 
angles. 


DIVISION  OF  THE  CIRCUMFERENCE. 

I.  For  the  purposes  of  trigonometrical  calculation,  the  cir- 
cumference of  the  circle  is  divided  into  360  equal  parts,  called 
degrees ;  each  degree  into  60  equal  parts,  called  minutes ;  and 
each  minute  into  60  equal  parts,  called  seconds. 

The  semicircumference,  or  the  measure  of  two  right  angles, 
contains  180  degrees  ;  the  quarter  of  the  circumference,  usually 
denominated  the  quadrant,  and  which  measiu'es  the  right  an- 
gle, contains  90  degrees. 

II.  Degrees,  minutes,  and  seconds,  are  respectively  desig- 


214 


PLANE  TRIGONOMETRY. 


nated  by  the  characters  ;  «, ',  " :  thus  the  expression  16°  6'  15" 
represents  an  arc,  or  an  angle,  of  16  degrees,  6  minutes,  and 
15  seconds. 

III.  The  complememt  of  an  angle,  or  of  an  arc,  is  what  re- 
mains after  taking  that  angle  or  that  arc  from  90°.  Thus  the 
complement  of  25°  40' is  equal  to  90°— 25°  40' r=  64°  20' ;  and 
the  complement  of  12°  4'  32"  is  equal  to  90°— 12°  4'  32"  =  77° 
55'  28". 

In  general,  A  being  any  angle  or  any  arc,  90° — A  is  the  com^ 
plement  of  that  angle  or  arc.  If  any  arc  or  angle  be  added 
to  its  complement,  the  sum  will  be  90°.  Whence  it  is  evident 
that  if  the  angle  or  arc  is  greater  than  90°,  its  complement  will 
be  negative.  Thus,  the  complement  of  160°  34'  10"  is  — 70° 
34'  10".  In  this  case,  the  complement,  taken  positively,  would 
be  9,  quantity,  which  being  subtracted  from  the  given  angle  or 
arc,  the  remainder  would  be  equal  to  90°. 

The  two  acute  angles  of  a  right-angled  triangle,  are  together 
equal  to  a  right  angle  ;  they  are,  therefore,  complements  of  each 
other. 

IV.  The  supplement  of  an  angle,  or  of  an  arc,  is  what  re- 
mains after  taking  that  angle  or  arc  from  180°.  Thus  A  being 
any  angle  or  arc,  180° — A  is  its  supplement. 

In  any  triangle,  either  angle  is  the  supplement  of  the  sum  of 
the  two  others,  since  the  three  together  make  180°. 

If  any  arc  or  angle  be  added  to  its  supplement,  the  sum  will 
be  180°.  Hence  if  an  arc  or  angle  be  greater  thaii  180°,  its 
supplement  will  be  negative.  Thus,  the  supplement  of  200° 
is  — 20°.  The  supplement  of  any  angle  of  a  triangle,  or  indeed 
of  the  sum  of  either  two  angles,  is  always  positive, 


GENERAL  IDEAS  RELATING  TO  TRIGONOMETRICAL  LINES. 


V.  The  sine  of  an  arc  is 
the  perpendicular  let  fall  from 
one  extremity  of  the  arc,  on 
the  diameter  which  passes 
through  the  other  extremity. 
Thus,  MP  is  the^ine  of  the 
arc  AM,  or  of  the  angle  ACM. 

The  tangent  of  an  arc  is  a 
line  touching  the  arc  at  one 
extremity/ and  limited  by  the 
prolongation  of  the  diameter 
which  passes  through  the 
other  extremity.'.  Thus  AT  is 
the  tangent  of  tne  arc  AM, 
or  of  the  angle  ACM.  - 


PLANE  TRIGONOMETRY.  215 

The  secant  of  an  arc  is  the  line  drawn  from  the  centre  of 
the  circle  through  one  extremity  of  the  arc  land  limited  by  the 
tangent  drawn  through  the  other  extremity.  Thus  CT  is  the 
secant  of  the  arc  AM,  or  of  the  angle  ACM. 

The  versed  sine  of  an  arc,  is  the  part  of  the  diameter  inter- 
cepted between  one  extremity  of  the  arc  and  the  foot  of  the 
sine.  Thus,  AP  is  the  versed  sine  of  the  arc  AM,  or  the  angle 
ACM. 

These  four  lines  MP,  AT,  CT,  AP,  are  dependent  upon  the 
arc  AM,  and  are  always  determined  by  it  and  the  radius ;  they 
are  thus  designated : 

MP=sin  AM,  or  sin  ACM, 
ATintangAM,  or  tang  ACM, 
CTz^secAM,  or  sec  ACM, 
APr=ver-sin  AM,  or  ver-sin  ACM. 
VI.  Having  taken  the  arc  AD  equal  to  a  quadrant,  from  the 
points  M  and  D  draw  the  lines  MQ,  DS,  perpendicular  to  the 
radius  CD,  the  one  terminated  by  that  radius,  the  other  termi- 
nated by  the  radius  CM  produced  ;  the  lines  MQ,  DS,  and  CS, 
will,  in  like  manner,  be  the  sine,  tangent,  and  secant  of  the  arc 
MD,  the  complement  of  AM.     For  the  sake  of  brevity,  they 
are  called  the  cosine,  cotangent,  and  cosecant,  of  the  arc  AM, 
and  are  thus  designated  : 

MQ=cosAM,  or  cos  ACM, 
DS=cot  AM,  or  cot  ACM, 
I  CS=cosec  AM,  or  cosec  ACM. 

In  general,  A  being  any  arc  or  angle,  we  have 
cos  A=sin  (90°— A), 
cot  A = tang  (90°— A), 
cosec  A  =  sec  (90° — A). 
The  triangle  MQC  is,  by  construction,  equal  to  the  triangle 
CPM  ;  consequently  CPrrrMQ  :  hence  in  the  right-angled  tri- 
angle CMP,  whose  hypothenuse  is  equal  to  the  radius,  the  two 
sides  MP,  CP  are  the  sine  and  cosine  of  the  arc  AM :  hence, 
the  cosine  of  an  arc  is  equal  to  that  part  of  the  radius  inter- 
cepted between  the  centre  and  foot  of  the  sine. 

The  triangles  CAT,  CDS,  are  similar  to  the  equal  triangles 
CPM,  CQM ;  hence  they  are  similar  to  each  other.  From 
these  principles,  we  shall  very  soon  deduce  the  different  rela- 
tions  which  exist  between  the  lines  now  defined  :  before  doing 
so,  however,  we  must  examine  the  changes  which  those  lines 
undergo,  when  the  arc  to  which  they  relate  increases  from  zero 
to  180«. 

The  angle  ACD  is  called  the  first  quadrant ;  the  angle  DCB, 
the  second  quadrant ;  the  angle  BCE,  the  third  quadrant ;  and 
the  angle  EC  A,  the  fourth  quadrant. 


210 


PLANE  TRIGONOMETRY. 


B 


D 

N 

^ 

Q^^ 

V 

\ 

P^ 

/ 

p'   \ 

y 

I 

\ 

k 

T^' 

R       J 

s 


E 


VII.  Suppose  one  extrem- 
ify  of  the  arc  remains  fixed  in 
A,  wiiile  the  other  extremit}% 
marked  M,  runs  successively 
throughout  the  whole  extent 
of  the  semicircumference, 
from  A  to  B  in  the  direction 
ADB. 

When  the  point  M  is  at  A, 
or  when  the  arc  AM  is  zero, 
the  three  points  T,  M,  P,  are 
confounded  with  the  point  A ; 
whence  it  appears  that  the 
sine  and  tangent  of  an  arc 

zero,  are  zero,  and  the  cosine  and  secant  of  this  same  arc,  are 
each  equal  to  the  radius.     Hence  if  R  represents  the  radius  ol 
the  circle,  we  have 
V  sin  0=0,  tang  0  —  0,  cos  0=R,  secO=R. 

VIII.  As  the  point  M  advances  towards  D,  the  sine  increases, 
and  so  likewise  does  the  tangent  and  the  secant;  but  the  cosine, 
the  cotangent,  and  the  cosecant,  diminish. 

When  the  point  M  is  at  the  middle  of  AD,  or  when  the  arc 
AM  is  45°,  in  which  case  it  is  equal  to  its  complement  MD, 
the  sine  MP  is  equal  to  the  cosine  MQ  or  CP  ;  and  the  trian- 
gle CMP,  having  become  isosceles,  gives  the  proportion 
MP  :  CM  :  :  1  :   x/2, 
or  sin  45°  :  R  :  :  1  :   V2. 


Hence 


sin  45°=cos45o=-7-=:iR\/2 
V  2 


In  this  same  case,  the  triangle  CAT  becomes  isosceles  and 
equal  to  the  triangle  CDS  ;  whence  the  tangent  of  45°  and  its 
cotangent,  are  each  equal  to  the  radius,  and  consequently  we 
have 

tang  45°  =  cut  ..15° =R. 

IX.  The  arc  AM  continuing  to  increase,  the  sine  increases 
till  M  arrives  at  D  ;  at  which  point  the  si^ie  is  equal  to  the  ra- 
dius, and  the  cosine  is^zero.     Hence  we  have   , 

sin90°=:R,    cos  90°  =  0; 

and  it  may  be  observed,  that  these  values  are  a  consequence 

of  the  values  already  found  for  the  sine  and  cosine  of  the  aiv 

zero  ;  because  the  complement  of  90«  being  zero,  we  have 

sin  yO''— cos  0°=R,  and 

cos  90°=rsin  0°=0. 


PLANE  TRIGONOMETRY.  211 

As  to  the  tangent,  it  increases  very  rapidly  as  the  point  M 
approaches  D  ;  and  finally  when  this  point  reaches  D,  the  tan- 
gent properly  exists  no  longer,  because  the  lines  AT,  CD, 
being  parallel,  cannot  meet.  This  is  expressed  by  saying  that 
the  tangent  of  90°  is  infinite  ;  and  we  write  tang  90" r:  ao 
The  complement  of  90"  being  zero,  we  have 

tang  O=cot  90"  and  cot  Orztang  90°. 

Hence       cot  90° =0,  and  cot  0=ao  . 

X.  The  point  M  continuing  to  advance  from  D  towards  B, 
the  sines  diminish  and  the  cosines  increase.  Thus  MT'  is  the 
sine  of  the  arc  AM',  and  M'Q,  or  CP'  its  cosine.  But  the  arc 
M'B  is  the  supplement  of  AM',  since  AM'  +  M'B  is  equal  to  a 
semicircumference  ;  besides,  if  M'M'  is  drawn  parallel  to  AB, 
the  arcs  AM,  BM',  which  are  included  between  parallels,  will 
evidently  be  equal,  and  likewise  the  perpendiculars  or  sines 
MP,  M'P'.  Hence/i/ie  sine  of  an  arc  or  of  an  angle  is  equal  to 
the  sine  of  the  siipplement  of  that  arc  or  angle^ 

The  arc  or  angle  A  has  for  its  supplement  180" — A:  hence 
generally,  we  have 

sin  A = sin  (180"— A.) 
The  same  property  might  also  be  expressed  by  the  equation 

sin  (90^  +  B)  =  sin  (90°— B), 
B  being  the  arc  DM  or  its  equal  DM'. 

XI.  The  same  arcs  AM,  AM',  which  are  supplements  of 
each  other,  and  which  have  equal  sines,  have  also  equal  co- 
sines CP,  CP' ;  but  it  must  be  observed,  that  these  cosines  lie 
in  different  directions.  The  line  CP  which  is  the  cosine  of  the 
arc  AM,  has  the  origin  of  its  value  at  the  centre  C,  and  is  esti- 
mated in  the  direction  from  C  towards  A ;  while  CP',  the  cosine 
of  AM'  has  also  the  origin  of  its  value  at  C,  but  is  estimated  in 
a  contrary  direction,  from  C  towards  B. 

Some  notation  must  obviously  be  adopted  to  distinguish  the 
one  of  such  equal  lines  from  the  other  ;  and  that  they  may  both 
be  expressed  analytically,  and  in  the  same  general  formula,  it  is 
necessary  to  consider  all  lines  which  are  estimated  in  one  di- 
rection as  j)ositiv)%  and  those  which  are  estimated  in  the  con- 
trary direction  as  negative.  If,  therefore,  the  cosines  which 
are  estimated  from  C  towards  A  be  considered  as  positive, 
those  estimated  from  C  towards  B,  must  be  regarded  as  nega- 
tive.    Hence,  generally,  we  shall  have, 

cos  A= — cos  (180° — A) 
that  is,frAe  cosine  of  an  arc  or  angle  is  equal  to  the  cosine  of  its 
supplement  taken  negatively^ 

The  necessity  of  changing  the  algebraic  sign  to  correspond 

T 


218 


PLANE  TRIGONOMETRY 


with  the  change  of  direction 
in  the  trigonometrical  line, 
may  be  illustrated  by  the  fol- 
lowing example.  The  versed 
sine  AP  is  equal  to  the  radius 
CA  minus  CP  the  cosine  AM  : 
that  is, 

ver-sin  AM.=:R— cos  AM. 
Now  when  the  arc  AM  be- 
comes AM'  the  versed  sine 
AP,  becomes  AF,  that  is  equal 
to  R  +  CP'.  But  this  expression 
cannot  be  derived  from  the 
formula, 

ver-sin  AM=r:R — cos  AMj 
unless  we  suppose  the  cosine  AM  to  become  negative  as  soon 
as  the  arc  AM  becomes  greater  than  a  quadrant. 

At  the  point  B  the  cosine  becomes  equal  to  — R ;  that  is, 
cos  180^=— R. 

For  all  arcs,  such  as  ADBN',  which  terminate  in  the  third 
quadrant,  the  cosine  is  estimated  from  C  towards  B,  and  is 
consequently  negative.  At  E  the  cosine  becomes  zero,  and  for 
all  arcs  which  terminate  in  the  fourth  quadrant  the  cosines  are 
estimated  from  C  towards  A,  and  are  consequently  positive. 

The  sines  of  all  the  arcs  which  terminate  in  the  first  and 
second  quadrants,  are  estimated  above  the  diameter  BA,  w^hile 
the  sines  of  those  arcs  which  terminate  in  the  third  and  fourth 
quadrants  are  estimated  below  it.  Hence,  considering  the 
former  as  positive,  we  must  regard  the  latter  as  negative. 

XII.  Let  us  now  see  what  sign  is  to  be  given  to  the  tangent 
of  an  arc.  The  tangent  of  the  arc  AM  falls  above  the  line  BA, 
and  w^e  have  already  regarded  the  lines  estimated  in  the  direc- 
tion At  as  positive  :  therefore  the  tangents  of  all  arcs  which 
terminate  in  the  first  quadrant  will  be  positive.  But  the  tan- 
gent of  the  arc  AM',  greater  than  90^,  is  determined  by  the 
intersection  of  the  two  lines  M'C  and  AT.  These  lines,  how- 
ever, do  not  meet  in  the  direction  AT  ;  but  they  meet  in  the 
opposite  direction  AV.  But  since  the  tangents  estimated  in  the 
direction  AT  are  positive,  those  estimated  in  thfe  direction  AV 
must  be  negative  :  therefore,  the  tangents  of  all  arcs  which  ter- 
minate in  the  second  quadrant  will  be  negative. 

When  the  point  M'  reaches  the  point  B  the  tangent  AV  will 
become  equal  to  zero :  that  is, 

tang  1 80°  =  0. 

When  the  point  M'  passes  the  point  B,  and  comes  into  the 
position  N',  the  tangent  of  the  arc  ADN'  will  be  the  line  AT : 


PLAJSE  TRIGONOMETRY.  219 

hence,^  the  tangents  of  all  arcs  which  terminate  in  the  third  quad- 
rant are  positive. 

At  E  the  tangent  becomes  infinite :  that  is, 
tang270°  =  Q0. 

When  the  point  has  passed  along  into  the  fourth  quadrant 
to  N,  the  tangent  of  the  arc  ADN'N  will  be  tl^e  line  AV :  hence, 
the  tangents  of  all  arcs  which  terminate  in  the  fourth  quadrant 
are  negative. 

The  cotangents  are  estimated  from  the  line  ED.  Those  which 
lie  on  the  side  DS  are  regarded  as  positive,  and  those  which  lie 
on  the  side  DS'  as  negative.  /"Hence,  the  cotangents  are  posi- 
tive in  the  first  quadrant,  negative  in  the  second,  positive  in  the 
third,  and  negative  in  the  fourth.\  When  the  point  M  is  at  B 
the  cotangent  is  infinite  ;  when  m  E  it  is  zero :  hence, 

cot  180°=— 00  ;  cot  270°  =  0. 
Let  q  stand  for  a  quadrant ;  then  the  following  table  will  show 
the  signs  of  the  trigonometrical  lines  in  the  different  quadrants. 

Iq  2q  Sq  4q 

Sine  +  +  •—  — 

Cosine  +  —  —  +       . 

Tangent  +  —  +  — 

Cotangent  4-  —  -{-  — 

XIII.  In  trigonometry,  the  sines,  cosines,  iSz^c.  of  arcs  or  an- 
gles greater  than  180°  do  not  require  to  be  considered  ;  the 
angles  of  triangles,  rectilineal  as  well  as  spherical,  and  the 
sides  of  the  latter,  being  always  comprehended  between  0  and 
180°.  But  in  various  applications  of  trigonometry,  there  is  fre- 
quently occasion  to  reason  about  arcs  greater  than  the  semi- 
circumference,  and  even  about  arcs  containing  several  circum- 
ferences. It  will  therefore  be  necessary  to  find  the  expression 
of  the  sines  and  cosines  of  those  arcs  whatever  be  their 
magnitude. 

We  generally  consider  the  arcs  as  positive  which  are  esti- 
mated from  A  in  the  direction  ADB,  and  then  those  arcs  must 
be  regarded  as  negative  which  are  estimated  in  the  contrary 
dii-ection  AEB. 

We  observe,  in  the  first  place,  that  two  equal  arcs  AM,  AN 
with  contrary  algebraic  signs,  have  equal  sines  MP,  PN,  with 
contrary  algebraic  signs ;  while  the  cosine  CP  is  the  same  for 
both. 

The  equal  tangents  AT,  AV,  as  well  as  the  equal  cotangents 
DS,  DS',  have  also  contrary  algebraic  signs.  Hence,  calling 
X  the  arc,  we  have  in  general, 

sin  {^x)= — sin  x 
'  i  cos  ( — x)  =  cos  X  \ 

\  tang  ( — x)  = — tang  x     \ 
cot  ( — x)= — cot  a; 


220 


PLANE  TRIGONOMETRY. 


By  considering  the  arc  AM,  and  its  supplement  AM',  and 
recollecting  what  has  been  said,  we  readily  see  that, 
sin  (an  arc)  =  sin  (its  supplement) 
cos  (an  arc)  = — cos  (its  supplement) 
tang  (an  arc)  :=— tang  (its  supplement) 
cot  (an  arc)  = — cot  (its  supplement). 

It  is  no  less  evident,  that      e'  D  ^ 

if  one  or  several  circumfe- 
rences were  added  to  any 
arc  AM,  it  would  still  termi- 
nate exactly  at  the  point  M, 
and  the  arc  thus  increased 
would  have  the  same  sine  as 
the  arc  AM ;  hence  if  C  rep- 
resent a  w^hole  circumfe- 
rence or  360°,  we  shall  have 
sin  X = sin  (C  +  x)=  sin  x = sin 
(2C  +  x),  &c. 

The  same  observation  is  ap- 
plicable to  the  cosine,  tan- 
gent, &c. 

Hence  it  appears,  that  whatever  be  the  magnitude  of  x  the 
proposed  arc,  its  sine  may  always  be  expressed,  with  a  proper 
sign,  by  the  sine  of  an  arc  less  than  180°.  For,  in  the  first 
place,  we  may  subtract  360°  from  the  arc  x  as  often  as  they 
are  contained  in  it ;  and  y  being  the  remainder,  we  shall  have 
sin  a;=sin  y.  Then  if?/  is  greater  than  180°,  make  y— 180°-f-2, 
and  we  have  sin  y= — sin  z.  Thus  all  the  cases  are  reduced 
to  that  in  which  the  proposed  arc  is  less  than  180°  ;  and  since 
we  farther  have  sin  (90°  +  a;)  =  sin  (90° — x),  they  are  likewise 
ultimately  reducible  to  the  case,  in  which  the  proposed  arc  is 
between  zero  and  90°. 

XIV.  The  cosines  are  always  reducible  to  sines,  by  means 
of  the  formula/ cos  A  =  sin  (90° — A)J  or  if  we  require  it,  by 
means  of  the  fArmula  cos  A  =  sin  (90° -f  A) :  and  thus,  if  we  can 
find  the  value  of  the  sines  in  all  possible  cases,  we  can  also  find 
that  of  the  cosines.  Besides,  as  has  already  been  shown,  that 
the  negative  cosines  are  separated  from  the  posilive  cosines  by 
the  diameter  DE;  all  the  arcs  whose  extremities  fall  on  the 
right  side  of  DE,  having  a  positive  cosine,  while  those  whose 
extremities  fall  on  the  left   have  a  negative  cosine. 

Thus  from  0°  to  90°  the  cosines  are  positive ;  from  90°  to 
270°  they  are  negative  ;  from  270°  to  300°  they  again  become 
positive ;  and  after  a  whole  revolution  they  assume  the  same 
values  as  in  the  preceding  revolution,  for  cos  (360°+a;)=cosa:. 


PLANE  TRIGONOMETRY. 


221 


From  these  explanations,  it  will  evidently  appear,  that  the 
sines  and  cosines  of  the  various  arcs  which  are  multiples  of  the 
quadrant  have  the  following  values : 


sin      0^  =  0 

sin    90°=R 

cos      0°=R 

cos    90°  =0 

sin  180°=0 

sin  270°=— R 

cos  180°=— R 

cos  270° =0 

sin  360^=0 

sin  450° =R 

cos  360° =R 

cos  450° =0 

sin  540° =0 

sin  630°=— R 

cos  540°=— R 

cos  630°=0 

sin  720° =0 

sin  810°=R 

cos  720° =R 

cos  810°=0 

&c. 

&c. 

&c. 

&c. 

And  generally,  k  designating  any  whole  number  we  shall 
have 

sin  2A;.90°=0,  cos  (2A:+1)  .  90°  =  0, 

sin  (4A;+  1)  .  90°=R,  cos  Ak  .  90°=R, 

sin  (4^—1)  .  90°=— R,  cos  (4A;  +  2)  .  90°=— R. 
What  we  have  just  said  concerning  the  sines  and  cosines 
renders  it  unnecessary  for  us  to  enter  into  any  particular  de- 
tail respecting  the  tangents,  cotangents,  &c.  of  arcs  greater 
than  180°  ;  the  value  of  these  quantities  are  always  easily  de- 
duced from  those  of  the  sines  and  cosines  of  the  same  arcs  : 
as  we  shall  see  by  the  formulas,  which  we  now  proceed  to 
explain. 


THEOREMS   AND  FORMULAS  RELATING  TO  SINES,  COSINES, 
TANGENTS,  &c. 


\ 


XV.^T/ie  sine  of  an  arc  is  half  the  chord  which  subtends  a 
^  double  arc,^ 


'■) 


For  the  radius  CA,  perpen- 
dicular to  the  chord  MN,  bi- 
sects this  chord,  and  likewise 
the  arc  MAN ;  hence  MP,  the 
sine  of  the  arc  MA,  is  half  the 
chord  MN  which  subtends 
the  arc  MAN,  the  double  of 
MA. 

The  chord  which  subtends 
the  sixth  part  of  the  circum- 
ference is  equal  to  the  radius  ; 
hence 

^^^'orsin30°=iR,) 


sm 


12 


in  other  words,  the  sine  of  a  third  part  of  the  right  angle  is 
equal  to  the  half  of  the  radius. 


T  * 


222 


PLANE  TIIIGOIMOMETRY. 


X( 


I  v^ 


^  XVI.  The  square  of  the  sine 
hf  an  arc,  together  with  the 
Square  of  the  cosine,  is  equal 
;io  the  square  of  the  radius  ;  so 
that  in  general  terms  we  have 
sin^A  +  cos2A=:R2. 

This  property  results  im- 
mediately from  the  right-an- 
gled triangle  CMP,  in  which 
MP-+CP'^=CM2. 

It  follows  that  when  the 
sine  of  an  arc  is  given,  its  co- 
sine may  be  found,  and  re- 
ciprocally, by  means  of  the 

formulas  cos  A  =  d=  V  (R^ — sin^A),  and  sin  A  =  rfc  \/  (R^ — cos^A). 
The  sign  of  these  formulas  is  +,  or  — ,  because  the  same  sine 
MP  answers  to  the  two  arcs  AM,  AM',  whose  cosines  CP,  CP', 
are  equal  and  have  contrary  signs ;  and  the  same  cosine  CP 
answers  to  the  two  arcs  AM,  AN,  whose  sines  MP,  PN,  are 
also  equal,  and  have  contrary  signs. 

Thus,  for  example,  having  found  sin  30^=|R,  we  may  de- 
duce from  itcos30^orsin60^=\/(R2— iR2)  =  yfR2— iRv/3, 


XVII.  The  sine  and  cosine  of  an  drc  A'heing  'given,  it  is  re-^ 
quired  to  find  the  tangent,  secant,  cotangent,  and  cosecant  of  the 
same  arc. 

The  triangles  CPM,  CAT,  CDS,  being  similar,  we  have  the 
proportions  : 


CP  :  PM  :  :  CA  :  AT ;  or  cos  A  :  sin  A  : :  R  :  tang  A= 


CP  :  CM  :  :  CA  :  CT  ;  or  cos  A  :  R  :  :  R  :  sec  A  = 


PM  :  CP  : :  CD  :  DS  ;  or  sin  A  :  cos  A  :  :  R  :  cot  A= 


R  sin  A 
cos  A 

_R2_ 
cos  A 

RcosA 


PM  :  CM  :  :  CD  :  CS ;  or  sin  A  :  R  : :  R  :  cosec  A=- 


sin  A 
R^ 

sin  A 


which  are  the  four  formulas  required.  It  may  also  be  observed, 
that  the  two  last  formulas  might  be  deduced  from  the  first  two, 
by  simply  putting  90° — A  instead  of  A. 

From  these  formulas,  may  be  deduced  the  values,  with  their 
proper  signs,  of  the  tangents,  secants,  &c.  belonging  to  any 
arc  whose  sine  and  cosine  are  known  ;  and  since  the  progres- 
sive law  of  the  sines  and  cosines,  according  to  the  different 
arcs  to  which  they  relate,  has  been  developed  already,  it  is 
unnecessary  to  say  more  of  the  law  which  regulates  the  tan- 
gents and  secants.     ,- 


PLANE  TRIGONOMETRY.  223 

By  means  of  these  formulas,  several  results,  which  have 
already  been  obtained  concerning  the  trigonometrical  lines, 
may  be  confirmed.  If,  for  example,  we  make  A  =00^,  we 
shall  have  sin  A=R,  cos  A  — 0  ;  and  consequently  tang  90°-= 

W 

— ,  an  expression  which  designates  an  infinite  quantity  ;  for 

the  quotient  of  radius  divided  by  a  very  small  quantity,  is  very 
great,  and  increases  as  the  divisor  diminishes  ;  hence,  the  quo- 
tient of  the  radius  divided  by  zero  is  greater  than  any  finite 
quantity. 

The  tangent  being  equal  to  R ;  and  cotangent  to  R.-^-  ; 

cos  siii. 


.V 

itf( 


follows  that  tangent  and  cotangent  will  both  be  positive 
when  the  sine  and  cosine  have  like  algebraic  signs,  and  both 
negative,  when  the  sine  and  cosine  have  contrary  algebraic 
signs.  Hence,  the  tangent  and  cotangent  have  the  same 
sign  in  the  diagonal  qiiadrants  :  that  is,  positive  in  the  1st  and 
3d,  and  negative  in  the  2d  and  4th  ;  results  agreeing  witii  those 
ofArt.  Xlf. 

The  Algebraic  signs  of  the  secants  and  cosecants  are  readily 
determined.  For,  the  secant  is  equal  to  radius  square  divided 
by  the  cosine,  and  since  radius  square  is  always  positive,  it 
follows  that  the  algebraic  sign  of  the  secant  will  depend  on 
that  of  the  cosine:  hence,  it  is  positive  in  the  1st  and  4th 
quadrants  and  negative  in  the  2nd  and  3id. 

Since  the  cosecant  is  equal  to  radius  square  divided  by  the 
sine,  it  follows  that  its  sign  will  depend  on  the  algebraic  sign 
of  the  sine  :  hence,  it  will  be  positive-  in  the  1st  and  2nd 
quadrants  and  negative  in  the  3rd  and  4th. 

XVIII.  The  formulas  of  the  preceding  Article,  combined 
with  each  other  and  with  the,  equation  sin  "A  +  cos ''^A:=R^ 
furnish  some  others  worthy  of  attention. 

First    we    have    R^  -f-  tang'^    A  =:  R-  +  "^IjI^-^  z=, 

cos^  A 

R-  (sin*^  A  +  cos-  A)         R''        ,  tiq  ,  *       2   a         •  o  * 

__1 i=: ;  hence  R^+tang^  A=sec-  A,  a 

cos  -A  cos"  A 

formula  which  might  be  immediately  deduced  from  the  righi- 
angled  triangle  CAT.  By  these  formulas,  or  by  the  right-an- 
gled triangle  CDS,  we  have  also  R'-^  +  cot- Arzcosec"  A. 

Lastly,  by  taking  the  product  of  the  two  formulas  tang  A=r 

RsinA 1 .    A     RcosA  ,  ,         .         .  *     tio 

--,  and  cot  A=  — —      we  have  tang  Ax  cot  A^R-,  a 

cos  A  sin  A 

R'  R2 

formula  which  gives  cot  A=  ^.^,^^   ^  ,  and  tang  A= 

We  likewise  have  cot  B= 


tang  A  '  o  ■  ■      cot  A, 


I 


tangB 


L.?-:. 


224 


PLANE  TRIGONOMETRY 


j  Hence  cot  A  :  cot  B  :  :  taiig  B  :  tang  A ;  that  is,  the  cotan- ' 
\  gents  of  two  arcs  are  reciprocally  proportional  to  their  tangents. 
The  formula  cot  Ax  tang  A=R'"^  might  be  deduced  imme- 
diately, by  comparing  the  similar  triangles  CAT,  CDS,  which 
give  AT  :  CA  :  :  CD  :  DS,  or  tang  A  :  R  :  :  R  :  cot  A 

XIX.  The  sines  and  cosines  of  two  arcs,  a  and  b,  being  given, 
it  is  required  to  find  the  sine  and  cosine  of  the  sum  or  difference 
of  these  arcs. 

Let  the  radius  AC=R,  the  arc 
AB=a,  the  arc  BD=6,  and  con- 
sequently ABD=a  +  h.  From 
the  points  B  and  D,  let  fall  the 
perpendiculars  BE,  DF  upon  AC ; 
from  the  point  D,  draw  DI  per- 
•  pendicular  to  BC  ;  lastly,  from 
the  point  I  draw  IK  perpendicu- 
lar, and  IL  parallel  to,  AC.  F'        C    FXTKE 

The  similar  triangles  BCE,  ICK,  give  the  proportions, 

sin  a  cos  /;. 


CB  :  CI  :  :  BE  :  IK,  or  R  :  cos  Z)  :  :  sin  «  :  IK= 


CB  :  CI  :  :  CE  :  CK,  or  R  :  cos  6  : :  cos  a  :  CK= 


R 

co§  a  cos '6. 


R 


The  triangles  DIL,  CBE,  having  their  sides  perpendicular, 
each  to  each,  are  similar,  and  give  the  proportions, 

CB  :  DI  :  :  CE  :  DL,  or  R  :  sin  ft  :  :  cos  a  :  DL= 


CB  :  DI  :  :  BE  :  IL,  or  R  :  sin  6  :  :  sin  a  :  IL: 


cos  a  sin  b, 

R 
sin  «  sin  b. 

R 


But  we  have 
IK+DL=DF: 

Hence 


sin  (a  +  6),  and  CK— IL-:CF=rcos  (a  +  b). 
sin  a  cos  ft  +  sin  i  cos  a 


sin  (a  +  b)  = 
cos  {a-\-b)  = 


R 

cos  a  cos  ft — sin  a*  sin  ft. 
R 


The  values  of  sin  (a — ft)  and  of  cos  {a — ft)  might  be  easily 
deduced  from  these  two  formulas;  but  they  may  be  found 
directly  by  the  same  figure.  For,  produce  the  sine  DI  till  it 
meets  the  circumference  at  M  ;  then  we  have  BM— BD=ft, 
and  MI==ID  =  sin  ft.  Through  the  point  M,  draw  MP  perpen- 
dicular, and  MN  parallel  to,  AC  ifsince  Mlrr^DI,  we  have  MN 
=IL,  and  IN=DD  But  we  have  IK— IN=MP  =  sin  («— ft), 
and  CK+MN=CP=^cos  (a—b)  ;  hence 


PLANE  TRIGONOMETRY.         '         225 

.     ,       ,.     sm  a  cos  b — sin  h  cos  a 
sm  {a—h)^ ^ 

-.      cos  a  cos  6  +  sin  a  sin  h 

cos  {a — o)  = — 

R 

These  are  the  formulas  which  it  was  required  to  find. 

The  preceding  demonstration  may  seem  defective  in  point 

of  generaHty,  since,  in  the  figure  which  we  have  followed,  the 

arcs  a  and  b,  and  even  a  +  &,  are  supposed  to  be  less  than  90°. 

But  first  the  demonstration  is  easily  extended  to  the  case  in 

which  a  and  b  being  less  than  90°,  their  sum  a  +  6  is  greater 

than  90°.     Then  the  point  F  would  fall  on  the  prolongation  of 

AC,  and  the  only  change  required  in  the  demonstration  would 

be  that  of  taking  cos  {a  -\-  b)  = — CF' ;  but  as  we  should,  at  the 

same  time,  have  CF'  =  rL' — CK',  it  would  still  follow  that  cos 

(a  +  b)  =  CK! — I'L',  or  R  cos  (a  +  b)=cos  a  cos  b — sin  a  sin  b. 

And  whatever  be  the  values  of  the  arcs  a  and  b,  it'  is  easily 

shown  that  the  formulas  arc  true :  hence  we  may  regard  them 

as  established  for  all  arcs.     We  will  repeat  and  number  the 

formulas  for  the  purpose  of  more  convenient  reference. 

.     ,    .  ,.      sin  a  cos  ^  +  sin  b  cos  a  .^  . 
sm  {a  +  b)  = g (1.). 

.     ,       ,v     sin  a  cos  b — sin  b  cos  a  .^. 
sin  (a—b)= ^^ (2.). 

,  ^     cos  a  cos  b — sin  a  sin  b  .    ^ 
cos  (a+b)=-. ^ (3.) 

,,     cos  a  cos  &  + sin  a  sin  ft  .. 
cos  (a—b)  = ^ (4.) 

XX.  If,  in  the  formulas  of  the  preceding  Article,  we  make 
6=a,  the  first  and  the  third  will  give     .      •■  ,  ^ 

.    _      2  sin  a  cos  a         ^       cos^  a — sin^  a     2  cos^  a — R" 
sm2a= p -f  cos  2a= ^ =-^=^ — ^ 

formulas  which  enable  us  to  find  the  sine  and  cosine  of  the 

double  arc,  when  we  know  the  sine  and  cosine  of  the  arc  itself. 

To  express  the  sin  a  and  cos  a  in  terms  of  ^a,  put  |a  for  a, 

and  we  have 

.    ^    2  sin  la  cos  ^a                     cos^  ^a — sin^  la 
sm  a= ± £_,     cos  a= i =— . 

R  R 

To  find  the  sine  and  cosine  of  Ja  in  terms  of  a,  take  the 
equations 

cos-  i^-hsin^  |«=R^  and  cos-i« — sin^  J-a=R  cos  a, 
there  results  by  adding  and  subtracting 

cos-  |a-^|R-  +  iR  cos  a,  and  sin2-i,Gt=iR2— iR  cos  a; 
whence 

sin  ia=  V  (|R2— iR  cos  a)=iV2R2— 2R  cos  «.' 
cos  'a=  \/(iR2+iR  cos  a)=iV2ilH2Rcosa. 


826  PLANE  TRIGONOMETRY. 

If  we  put  2a  in  the  place  of  a,  we  shall  have, 


sin  (?=  n/  (iR2_4R  cos  2a)  =i  V 2R"^— 2R  cos  2a. 

cos  a=V(iR2+iR  cos  2a)=\s/2W-^ 2R  cos  2a. 

Making,  in  the  two  last  formulas,  a=45°,  gives  cos2a=0,  and 

sin  45°=  \/iR2=RV| ;  and  also,  cos  45°=  VlR^R V-^. 

Next,  make  a =22°  30',  which  gives  cos  2a=R  V^,  and  we  have 

sin  22°  30'=R  V(i— ^Vi)  and  cos  22°  30'=RN/(^-+iVi). 

XXI.  If  we  multiply  together  formulas  (1.)  and  (2.)  Art. 
XIX.  and  substitute  for  cos'-^  a,  R^ — sin''^  a,  and  for  cos-  b, 
R^ — sin^  h ;  we  shall  obtain,  after  reducing  and  dividing  by  R^, 
sin  {a + h)  sin  {a — h) — sin^  a — sin^  h = (sin  a  +  sin  h)  (sin  a — sin  h), 

or,  sin  {a — V)  :  sin  a — sin  h  :  :  sin  a  +  sin  6  :  sin  (a +  6). 

XXII.  The  formulas  of  Art.  XIX.  furnish  a  great  number  of 
consequences  ;  among  which  it  will  be  enough  to  mention  those 
of  most  frequent  use.  By  adding  and  subtracting  we  obtain 
the  four  which  follow, 

2 
sm  («+6)  +  sin  (a — h)= — sin  a  cos  6. 

R 

2 

sin  {a-{-p) — sin  {a — 6)=-_sin  b  cos  a. 

R 

2 

cos  (a +  6)  + COS  (a — 6)  =  —- cos  a  cos  b. 
R 

2 

cos  {a — b) — cos  (a  +  6)=-psin  a  sin  b. 

and  which  serve  to  change  a  product  of  several  sines  or  co- 
sines into  linear  sines  or  cosines,  that  is,  into  sines  and  cosines 
multiplied  only  by  constant  quantities. 

XXIII.  If  in  these  formulas  we  put  a-{-b=pf  a — b=q,  which 

p-\-q        p — q 
gives  a=—^j  5=—^,  we  shall  find 

2 

sin  j9-f  sin  ^=-— -sin  ^(p  +  q)  cos  |(jo — q)  (1.) 

R       " 

2 
sin  p — sin  q~-:^sm  -J  (p — q)  cos  h{p  +  q)  (2.) 

R 

W  cosp  +  cosq  =  -^cos  -^  {p  +  q)  cos  h  (p — q)  (3.) 

cosg — cos  7?=^  sin  h  (p  +  q)  sin  J  {p — q)  (4.) 
R       " 


PLANE  TRIGONOMETRY.  227 

')  -  6^^"$^  /-  ■      ■ 

If  we  make  g=o,  we  shall  obtain, 

2  sin  i»  cos  \p 
sin  p  — ^-^ -— 

„                  2  cos^  \  p 
R  +  cos  p= „  " 

^  2  sin^  I  p 

R — cos  j9= :  hence 

R 

sin  j9     _  tang  ^p         R 

wky  R+cos /?~      R      ""cotjjp 

-  sinp     _^cot  ^p_      R    _ 

R — cos  p  R  tang  ^p- 

formulas  which  are  often  employed  in  trigonometrical  calcula- 
tions for  reducing  two  terms  to  a  single  one. 

XXIV.  From  the  first  four  formulas  of  Art  XXIII.  and  the  first 

of  Art.  XX.,dividinff,  and  considerinjsj  that =  — M—= 

^  ^         cos  a        R        cot  a 

we  derive  the  following : 

I  sin  ;7  +  sin  q  _^\n  \  {p -\- q)  cos  ^  (p—g^)  _  tang -|(;?  +  y) 

>   sin  j9 — sin  ^     cos|^  (^-f  g)  sin  J  (jo — q)     tangj(jt? — q) 

'    sin  ;?  +  sin  7_  sin  j^  (;?  +  9^)_tangl  (jp  +  y) 
cosp-fcos^'     Qos^{p-^q)  R 

i  s|g_jH;ig^^_cos^  (/?— y)  _cot  I  (p—q)  ^ 

'   cos  q — cos  p     sin  -J  (p — q)  R 

sin  /?— sin  q_s\nl- {p—q) _tanrri  (p—q) 
COSJ9  +  COS5'     COS  J  (2? — q)  R 

::in /> — sin  q  _<^os^(p  +  q)  ^cot^  (p  +  q) 
cos  q — COS  JO     sin  ^{p-\-q)  R 

( i)sp  +  cos  ^  _cos  }f  (p  +  q)  COS  |  {p — 9')_cot  h{p-rq) 
r  <s  q — cos  p  sin  h  (p+q)  sm  |-  (p — qj  tang-^  (p—q) 
sin  /)  +  sin  gf  2sin  j  (/?  +  ^)  cos  -}  (p — q)  cos  |  (jo — g) 
sin  (p  +  ^)~2sin  J  {p-^q)  cos  J-  (p  +  g)~cosi  (p  +  g) 
sin  p — sin  q  2sin  |  (p — g^)  cos  ^  (p  +  q)  sin  |^  (jo — q) 
sin     (io  +  ^)~2sini  (p  +  ^)cosi(;?  +  5')'^sin|  (;?  +  g') 

Formulas  which  are  the  expression  of  so  many  theorems. 
From  the  first,  it  follows  that  the  sum  of  the  sines  of  two  arcs  is 
to  the  difference  of  these  sines,  as  the  tangent  of  half  the  sum  of 
\    the  arcs  is  to  the  tangent  of  half  their  difference. 


828  PLANE  TRIGONOMETRY. 

XXV.  In  order  likewise  to  develop  some  formulas  relative 
to  tangents,  let  us  consider  the  expression 

.    ,  ,.     R  sin  («  +  &)   .       ,  .  ,   ,         ,     .     .        , 

tang  {a-\-o)= -^ — — i,  m  which  by  suostitutmg  the  values 

^      cos  (a^b)  ''  ^ 

of  sin  (a +6)  and  cos  {a+h),  we  shall  find 

*        /    .  i.\     R  (sin  a  cos  &  +  sin  6  cos  a) 

tang(cz  +  &)=— A ■  ^  .    ,    . \ 

cos  a  cos  0 — sm  bsma 

MT              ,           .          cos  a  tanff  a       ,    .    ,     cos  h  tang  h 
JNow  we  have  sm  a= p         ,  and  sm  &= ^ — ^--  : 

substitute  these  values,  dividing  all  the  terms  by  cos  a  cos  h  ; 
we  shall  have 

R^ — tang  a  tang  6 
which  is  the  value  of  the  tangent  of  the  sum  of  two  arcs,  ex- 
pressed by  the  tangents  of  each  of  these  arcs.  For  the  tangent 
of  their  difference,  we  should  in  like  manner  find 
R^  (tang  ^— tang  h) 
tang  {a^h)  =R2+tang  «  tang  6. 
Suppose  6=a ;  for  the  duplication  of  the  arcs,  we  shall  have 
the  formula 

2  B?  tang  a 

tang  2a—rfrr, — ~ : 

°  H~ — tang^a 

Suppose  &=2a ;  for  their  triplication,  we  shall  have  the  for- 
mula 

tang  3«=?^J&^5g_^±*^M^  ; 
*  W — tang  a  tang  2a' 

in  which,  substituting  the  value  of  tang  2  a,  we  shall  have 

3R^  tang  a — tang  ^a 

tang  3  a=  -_— — ^ & — 

^  R'^_3  tang  ^a.  ■» 

XXVI.  Scholium,  The  radius  R  being  entirely  arbitrary,  isl 
generally  taken  equal  to  1,  in  which  case  it  does  not  appear  in 
the  trigonometrical  formulas.     For  example  the  expression  for 
the  tangent  of  twice  an  arc  when  R=l,  becomes, 

2  tang  a 
tang2<z=  ^    - 


1 — tang-  fl- 
it we  have  an  analytical  formula  calculated  to  the  radius  of  1, 
and  wish  to  apply  it  to  another  circle  in  which  the  radius  is  R, 
we  must  multiply  each  term  by  such  a  power  of  R  as  will  make 
all  the  terms  homogeneous:  that  is,  so  that  each  shall  contain  the 
same  number  of  literal  factors. 


PLANE  TRIGONOMETRY.  229 


CONSTRUCTION  AND  DESCRIPTION  OF  THE  TABLES. 

XXVII.  If  the  radius  of  a  circle  is  taken  equal  to  1,  and  the 

lengths  of  the  lines  representing  the  sines,  cosines,  tangents, 

cotangents,  &c.  for  every  minute  of  the  quadrant  be  calculated, 

and  written  in  a  table,  this  would  be  a  table  of  natural  sines, 

..cosines,  &c.  sj^ 

W^   XXVIII.    If  such  a  table  were  known,  it  would  be  easy  to 
^  calculate  a  table  of  sines,  &c.  to  any  other  radius ;  since,  in 
ditferent  circles,  the  sines,  cosines,  &c.  of  arcs  containing  the 
same  number  of  degrees,  are  to  each  other  as  their  radii. 

XXIX.  If  the  trigonometrical  lines  themselves  were  used,  it 
would  be  necessary,  in  the  calculations,  to.  perform  the  opera- 
tions of  multiplication  and  division.  To  avoid  so  tedious  a 
method  of  calculation,  we  use  the  logarithms  of  the  sines,  co- 

•  sines,  &:c. ;  so  that  the  tables  in  common  use  show  the  values 
of  the  logarithms  of  the  sines,  cosines,  tangents,  cotangents,  &c. 
for  each  degree  and  minute  of  the  quadrant,  calculated  to  a 

-  given  radius.  This  radius  is  10,000,000,000,  and  consequently 
its  logarithm  is  10. 

XXX.  Let  us  glance  for  a  moment  at  one  of  the  methods 
of  calculating  a  table  of  natural  sines. 

The  radius  of  a  circle  being  1,  the  semi-circumference  is  known 
to  be  3.14159265358979.  This  being  divided  successively,  by 
180  and  60,  or  at  once  by  10800,  gives  .0002908882086657, 
for  the  arc  of  1  minute.  Of  so  small  an  arc  the  sine,  chord, 
and  arc,  differ  almost  imperceptibly  from  the  ratio  of  equality  ; 
so  that  the  first  ten  of  the  preceding  figures,  that  is,  .0002908882 
may  be  regarded  as  the  sine  of  1' ;  and  in  fact  the  sine  given 
in  the  tables  which  run  to  seven  places  of  figures  is  .0002909. 
By  Art.  XVI.  we  have  for  any  arc,  cos=  V(l — sin^).  This 
theorem  gives,  in  the  present  case,  cos  1'  =  .9999999577.  Then 
by  Art.  XXII.  we  shall  haVe 

2  cos  I'xsin  1'-— sin  0'=sin  2' =  .00058 17764 
2  cos  I'xsin  2'— sin  r=sin  3' =.0008726646 
2  cos  I'xsin  3'~-sin  2'  =  sin  4' =  .00 11635526 
2  cos  I'xsin  4'— sin  3'  =  sin  5'  =  .0014544407  * 
2  cos  I'xsin  5'— sin  4'=sin  6' =  .0017453284 
&c.  &c.  &c. 

Thus  may  the  work  be  continued  to  any  extent,  the  whole 
difficulty  consisting  in  the  multiplication  of  each  successive  re- 
sult by  the  quantity  2  cos  1'— .1.9999999154. 

U 


230  PLANE  TRIGONOMETRY. 

Or,  the  sines  of  T  and  2'  being  determined,  the  work  might 
be  continued  thus  (Art.  XXI.) : 

sin  T:  sin  2' — sin  I' : :  sin  2'  +  sin  1' :  sin  3 
sin  2':  sin  3' — sin  1' : :  sin  3'  +  sin  1' :  sin  4' 
sin  3' :  sin  4' — sin  1' ::  sin  4'  + sin  1' :  sin  5' 
sin  4' :  sin  5' — sin  1' ::  sin  5'  + sin  1' :  sin  6' 
&c.  &;c.  &c. 

In  like  manner,  the  computer  might  proceed  for  the  sines  of 
degrees,  &c.  thus : 

sin  1°  :  sin  2°--sin  1°  : :  sin  2''  + sin  1°  :  sin  3° 
sin  2°  :  sin  3°-— sin  1°  : :  sin  3°  + sin  1°  :  sin  4° 
sin  3°  :  sin  4°— sin  1°  : :  sin  4°  + sin  1°  :  sin  5° 
&:c.  &c.  &c. 

Above  45°  the  process  may  be  considerably  simplified  by 
Ihe  theorem  for  the  tangents  of  the  sums  and  differences  of 
arcs.  For,  when  tiie  radius  is  unity,  the  tangent  of  45°  is  also 
unity,  and  tan  (a  +  b)  will  be  denoted  thus  : 

tan(45°  +  fe)=:l±i^. 
^  ^     1— tan  b 

And  this,  again,  may  be  still  further  simplified  in  practice 
The  secants  and  cosecants  may  be  found  from  the  cosines  and 
sines. 


TABLE  OF  LOGARITHMS. 

XXXI.  If  the  logarithms  of  all  the  numbers  between  1  and 
any  given  number,  be  calculated  and  arranged  in  a  tabular  form, 
such  table  is  called  a  table  of  logarithms.  The  table  annexed 
shows  the  logarithms  of  all  numbers  between  1  and  10,000. 

The  first  column,  on  the  left  of  each  page  of  the  table,  is  the 
column  of  numbers,  and  is  designated  by  the  letter  N ;  the  deci- 
mal part  of  the  logarithms  of  these  numbers  is  placed  directly 
opposite  them,  and  on  the  same  horizontal  line. 

The  characteristic  of  the  logarithm,  or  the  part  which  stands 
totheleftof  the  decimal  point,  is  always  known,  being  1  less  than 
the  places  of  integer  figures  in  the  given  number,  and  there- 
fore it  is  not  written  in  the  table  of  logarithms.  '  Thus,  for  all 
numbers  between  1  and  10,  the  characteristic  is  0:  for  num- 
bers between  10  and  100  it  is  1,  between  100  and  1000  it  is 
2,  &c. 


( 


\ 


PLANE  TRIGONOMETRY.  231 

*-  PROBLEM. 

To  find  from  the  table  the  logarithm  of  any  number, 

CASE  I. 

Wlien  the  number  is  less  than  100. 

Look  on  the  first  page  of  the  table  of  logarithms,  along  the 
columns  of  numbers  under  N,  until  the  number  is  found  ;  the 
number  directly  opposite  it,  in  the  column  designated  Log.,  is 
the  logarithm  sought. 

CASE  II. 
IVhen  the  number  is  greater  than  100,  and  less  than  10,000, 


Find,  in  the  column  of  numbers,  the  three  first  figures  of  the 
given  number.  Then,  pass  across  the  page,  in  a  horizontal 
line,  into  the  columns  marked  0,  1,2,  3,  4,  &c.,  until  you  come 
to  the  column  which  is  designated  by  the  fourth  figure  of  the 
given  number :  to  the  four  figures  so  found,  tw^o  figures  taken 
from  the  cohimn  marked  0,  are  to  be  prefixed.  If  the  four 
figures  found,  stand  opposite  to  a  row  of  six  figures  in  the  column 
marked  0,  the  two  figures  from  this  column,  which  are  to  be 
prefixed  to  the  four  before  found,  are  the  first  two  on  the  left 
hand  ;  but,  if  the  four  figures  stand  opposite  a  line  of  only  four 
figures,  you  are  then  to  ascend  the  column,  till  you  come  to  the 
line  of  six  figures :  the  two  figures  at  the  left  hand  are  to  be 
prefixed,  and  then  the  decimal  part  of  the  logarithm  is  obtained. 
To  this,  the  characteristic  of  the  logarithm  is  to  be  prefixed, 
which  is  always  one  less  than  the  places  of  integer  figures  in 
the  given  number.    Thus,  the  logarithm  of  1 122  is  3.049993. 

In  several  of  the  columns,  designated  0,  1,  2,  3,  &c.,  small 
dots  are  found.  Where  this  occurs,  a  cipher  must  be  written 
for  each  of  these  dots,  and  the  two  figures  which  are  to  be  prer 
fixed,  from  the  first  column,  are  then  found  in  the  horizontal 
line  directly  below.  Thus,  the  log.  of  2188  is  3.340047,  the  two 
dots  being  changed  into  two  ciphers,  and  the  34  from  the 
column  0,  prefixed.  The  two  figures  from  the  colum  0,  must 
also  be  taken  from  the  line  below,  if  any  dots  shall  have  been 
passed  over,  in  passing  along  the  horizontal  line  :  thus,  the  loga- 
rithm of  3098  is  3.491081,  the  49  from  the  column  0  being 
taken  from  the  line  310. 


232  PLANE  TRIGONOMETRY. 


CASE   III. 

When  the  number  exceeds  10,000,  or  consists  of  five  or  mort 
places  of  figures. 

Consider  all  the  figures  after  the  fourth  from  the  left  hand, 
as  ciphers.  Find,  from  the  table,  the  logarithm  of  the  first  four 
places,  and  prefix  a  characteristic  which  shall  be  one  less  than 
the  number  of  places  including  the  ciphers.  Take  from  the  last 
column  on  the  right  of  the  page,  marked  D,  the  number  on  the 
same  horizontal  line  with  the  logarithm,  and  multiply  this  num- 
ber by  the  numbers  that  have  been  considered  as  ciphers : 
then,  cut  off  from  the  right  hand  as  many  places  for  decimals 
as  there  are  figures  in  the  multiplier,  and  add  the  product,  so 
obtained,  to  the  first  logarithm :  this  sum  will  be  the  logarithm 
sought. 

Let  it  be  required  to  find  the  logarithm  of  672887.  The  log. 
of  672800  is  found,  on  the  1 1th  page  of  the  table,  to  be  5.827886, 
after  prefixing  the  characteristic  5.  The  corresponding  num- 
ber in  the  column  D  is  65,  which  being  multiplied  by  87,  the 
figures  regarded  as  ciphers,  gives  5655 ;  then,  pointing  ofif  two 
places  for  decimals,  the  number  to  be  added  is  56.55.  This 
number  being  added  to  5.827886,  gives  5.827942  for  the  loga- 
rithm of  672887 ;  the  decimal  part  .55,  being  omitted. 

This  method  of  finding  the  logarithms  of  numbers,  from  the 
table,  supposes  that  the  logarithms  are  proportional  to  their 
respective  numbers,  which  is  not  rigorously  true.  In  the  exam- 
ple, the  logarithm  of  672800  is  5.827886  ;  the  logarithm  of 
672900,  a  number  greater  by  100,  5.827951  :  the  difference  of 
the  logarithms  is  65.  Now,  as  100,  the  diiference  of  the  numbers, 
IS  to  65,  the  diflference  of  their  logarithms,  so  is  87,  the  diffe- 
rence between  the  given  number  and  the  least  of  the  numbers 
used,  to  the  difference  of  their  logarithms,  which  is  56.55 :  this 
diflference  being  added  to  5.827886,  the  logarithm  of  the  less 
number,  gives  5.827942  for  the  logarithm  of  672887.  The  use 
of  the  column  of  differences  is  therefore  manifest. 

When,  however,  the  decimal  part  which  is  to  be  omitted  ex- 
ceeds .5,  we  come  nearer  to  the  true  result  by  increasing  the 
next  figure  to  the  left  by  1  ;  and  this  will  be  done  in  all  the 
calculations  which  follow.  Thus,  the  difference  to  be  added, 
was  nearer  57  than  56 ;  hence  it  would  have  been  more  exact 
to  have  added  the  former  juimber. 

The  logarithm  of  a  vulgar  fraction  is  equal  to  the  loga- 
rithm of  the  numerator,  minus  the  logarithm  of  the  denom- 


I 


PLANE  TRIGONOMETRY.  233 

inator.     The  logarithm  of  a  decimal  fraction  is  found,  hy  ccn-X 
sidering  it  as  a  whole  numher,  and  then  prefixing  to  the  decimal  \ 
part  of  its  logarithm,  a  negative  characteristic,  greater  hy  unity   ' 
than  the  number  of  ciphers  between  the  decimal  point  and  the  first 
significant  place  of  figures.     Thus,  the  logarithm  of  .0412.  is 
2^614897.  • 


PROBLEM. 
To  find  from  the  table,  a  number  answering  to  a  given  logarithm, 

XXXII  Search,  in  the  column  of  logarithms,  for  the  decimal 
part  of  the  given  logarithm,  and  if  it  be  exactly  found,  set  down 
the  corresponding  number.  Then,  if  the  characteristic  of  the 
given  logarithm  be  positive,  point  otf,  from  the  left  of  the  number 
found,  one  place  more  for  whole  numbers  than  there  are  units 
in  the  characteristic  of  the  given  logarithm,  and  treat  the  other 
places  as  decimals ;  this  will  give  the  number  sought. 

If  the  characteristic  of  the  given  logarithm  be  0,  there  will 
be  one  place  of  whole  numbers  ;  if  it  be  — 1,  the  number  will 
be  entirely  decimal ;  if  it  be  — 2,  there  will  be  one  cipher  be- 
tween the  decimal  point  and  the  first  significant  figure  ;  if  it  be 
— 3,  there  will  be  two,  &c.  The  number  whose  logarithm  is 
1.492481  is  found  in  page  5,  and  is  31.08. 

But  if  the  decimal  part  of  the  logarithm  cannot  be  exactly 
found  in  the  table,  take  the  number  answering  to  the  nearest 
less  logarithm ;  take  also  from  the  table  the  corresponding  dif- 
ference in  the  column  D ;  then,  subtract  this  less  logarithm  from 
the  given  logarithm  ;  and  having  annexed  a  sufficient  number 
of  ciphers  to  the  remainder,  divide  it  by  the  difference  taken 
from  the  column  D,  and  annex  the  quotient  to  the  number  an- 
swering to  the  less  logarithm :  this  gives  the  required  number, 
nearly.  This  rule,  like  the  one  for  finding  the  logarithm  of  a 
number  when  the  places  exceed  four,  supposes  the  numbers  to 
be  proportional  to  their  corresponding  logarithms. 

Ex.  I.  Find  the  number  answering  to  the  logarithm  1.532708- 
Here, 

The  given  logarithm,  is  ...         1.532708 

Next  less  logarithm  of  34,09,  is       -         -         1.532627 

Their  difference  is  -         _         .         .       ~l        gj" 

And  the  tabular  difference  is  128  :  hence 
128)  81.00  (63 
which  being  annexed  to  34,09,  gives  34.0963  for  the  number 
answering  to  the  logarithm  1.532708. 


234        ,    .       PLANE  TRIGOKOMETRY, 

^  Ex.  2.     Required  the  ^number  answering  to  the  logarithnj 

3.233568. 

The  given  logarithm  is  3.233568 

The  next  less  tabular  logarithm  of  1712,  is  3.233504 

Diff.=  64 

Tab.  Difr.  =  253)  €A.OO-  (25 
Hence  the  number  sought  is  1712.25,  marking  four  places 
of  integers  for  the  characteristic  3. 


TABLE  OF  LOGARITHMIC  SINES. 

XXXIII.  In  this  table  are  arranged  the  logarithms  of  the 
numerical  values  of  the  sines,  cosines,  tangents,  and  cotangents, 
of  all  the  arcs  or  angles  of  the  quadrant,  divided  to  minutes, 
and  calculated  for  a  radius  of  10,000,000,000.  The  logarithm 
of  this  radius  is  10.  In  the  first  and  last  horizontal  line,  of  each 
page,  are  written  the  degrees  wiiose  logarithmic  sines,  &:c.  are 
expressed  on  the  page.  The  vertical  columns  on  the  left  and 
right,  are  columns  of  minutes. 


CASE  L 

To  findy  in  the  table,  the  logarithmic  sine,  cosine,  tangent,  or  co- 
tangent of  any  given  arc  or  angle. 

1.  If  the  angle  be  less  than  45*^,  look  in  the  first  horizontal 
Ime  of  the  difterent  pages,  until  the  number  of  degrees  be 
found  ;  then  descend  along  the  column  of  minutes,  on  the  left 
of  the  page,  till  you  reach  the  number  showing  the  minutes  ; 
then  pass  along  the  horizontal  line  till  you  come  into  the  column 
designated,  sine,  cosine,  tangent,  or  cotangent,  as  the  case  may 
be  :  the  number  so  indicated,  is  the  logarithm  sought.  Thus,  the 
sine,  cosine,  tangent,  and  cotangent  of  19""  55',  are  found  on 
page  37,  opposite  55,  and  are,  respectively,  9.532312,  9.973215, 
9.559097,  10.440903. 

2.  If  the  angle  be  greater  than  45°,  search  along  the  bottom 
fine  of  the  different  pages,  till  the  number  of  degrees  arc  found  ; 
then  ascend  along  the  column  of  minutes,  on  the  right  hand 
side  of  the  page,  till  you  reach  the  number  expressing  the  mi- 
nutes ;  then  pass  along  the  horizontal  line  into  the  columns 
designated  tang.,  cotang.,  sine,  cosine,  as  the  case  may  be  ;  the 
number  so  pointed  out  is  the  logarithm  required. 


PLANE  TRIGONOMETRY.  235 

It  will  be  seen,  that  the  column  designated  sine  at  the  top  of 
the  page,  is  designated  cosine  at  the  bottom  ;  the  one  desig- 
nated tang.,  by  cotang.,  and  the  one  designated  cotang.,  by 
tang. 

The  angle  found  by  taking  the  degrees  at  the  top  of  the  page, 
and  the  minutes  from  the  first  vertical  column  on  the  left,  is  the 
complement  of  the  angle,  found  by  taking  the  corresponding 
degrees  at  the*  bottom  of  the  page,  and  the  minutes  tiaced  up 
in  the  right  hand  column  to  the  same  horizontal  line.  This 
being  apparent,  the  reason  is  manifest,  why  the  columns  desig- 
nated sine,  cosine,  tang.,  and  cotang.,  when  the  degrees  are 
pointed  out  at  the  top  of  the  page,  and  the  minutes  countec* 
downwards,  ought  to  be  changed,  respectively,  into  cosine,  sine, 
cotang.,  and  tang.,  when  the  degrees  are  shown  at  the  bottom 
of  the  page,  and  the  minutes  counted  upwards. 

it'  the  angle  be  greater  than  90^,  we  have  only  to  subtract  it 
from  180^,  and  take  the  sine,  cosine,  tangent,  or  cotangent  of 
the  remainder. 

The  secants  and  cosecants  are  omitted  in  the  table,  being 
easily  found  from  the  cosines  and  sines. 

R2  . 

For,    sec.  = ;    or,  taking  the  logarithms,  log.  sec.=2 

COS. 

log.  R — log.  COS.— 20 — log.  COS. ;  that  is,  the  logarithmic  secant 
is  found  by  substracLing  the  logarithmic  cosine  from  20.     And 

R^ 

cosec.  — ,  or  log.  cosec.=2  log.  R — log.  sine  =  20 — log. 

sine 

sine  ;  that  is,  the  logarithmic  cosecant  is  found  by  subtracting  the 

logarithmic  sine  from  20. 

It  has  been  shown  that  R-rrtang.  x  cotang. ;  therefore,  2  log. 
Rr=:log.  tang.  +  log.  cotang.;  or  20=r:log.  tang.  +  log.  cotang. 

The  column  of  the  table,  next  to  the  column  of  sines,  and 
on  the  right  of  it,  is  designated  by  the  letter  D.  This  column 
is  calculated  in  the  following  manner.  Opening  the  table  at 
any  page,  as  42,  the  sine  of  24°  is  found  to  be  9.609313  ;  of 
24^  1',  9.609597 :  their  difference  is  284  ;  this  being  divided  by 
60,  the  number  of  seconds  in  a  minute,  gives  4.73,  which  is 
entered  in  the  column  I),  omitting  the  decimal  point.  Now, 
supposing  the  increase  of  the  logarithmic  sine  to  be  propor- 
tional to  the  increase  of  the  arc,  and  it  is  nearly  so  for  60",  it 
follows,  that  473  (the  last  two  places  being  regarded  as  deci- 
mals) is  the  increase  of  the  sine  for  1".  Similarly,  if  the  arc 
be  24°  20',  the  increase  of  the  sine  for  1",  is  465,  the  last  two 
places  being  decimals.  The  same  remarks  are  equally  appli- 
cable in  respect  of  the  column  D,  after  the  column  cosine,  and 
of  the  column  D,  between  the  tangents  and  cotangents.  The 
column  D,  between  the  tangents    and  cotangents,   answers 


23(5  PLANE  TRIGONOMETRY. 

to  either  of  these  columns ;  since  of  the  same  arc,  the  log 
tang. -f  log.  cotangiz:20.  Therefore,  having  two  arcs,  a  and  h, 
log.  tang  6  + log.  cotang  fe^log.  tang  fl  +  log.  cotang  a;  or, 
log.  tang  h — log.  tang  «r=log.  cotang  a — log.  cotang  h. 

Now,  if  it  were  required  to  find  the  logarithmic  sine  of  an 
arc  expressed  in  degrees,  minutes,  and  seconds,  we  have  only 
to  find  the  degrees  and  minutes  as  before  ;  then  multiply  the 
corresponding  tabular  number  by  the  seconds,  cut  off  two  places 
to  the  right  hand  for  decimals,  and  then  add  the  product  to  the 
number  first  found,  for  the  sine  of  the  given  arc.  Thus,  if  we 
wish  the  sine  of  40°  26'  28". 

The  sine  40°  26'         -         -         -         -         9.811952 

Tabular  difference  =:=  247 

Number  of  seconds  m    28 


Product=:69.16,  to  beadded     =  69.16 


Gives  for  the  sine  of  40°  26'  28"  =9.812021.16 

The  tangent  of  an  arc,  in  ,which  there  are  seconds,  is  found 
in  a  manner  entirely  similar.  In  regard  to  the  cosine  and  co- 
tangent, it  must  be  remembered,  that  they  increase  while  the 
arcs  decrease,  and  decrease  while  the  arcs  are  increased,  con- 
sequently, the  proportional  numbers  found  for  the  seconds  must 
be  subtracted,  not  added. 

Ex,  To  find  the  cosine  3°  40'  40". 

Cosine  3°  40'  9.999110 

Tabular  diflference  i=  13 

Number  of  seconds  =  40 


Product  m  5.20,  which  being  subtracted    =:    5.20 
Gives  for  the  cosine  of  3°  40'  40"      9.999104.80 


CASE  II. 

To  find  the  degrees,  minutes,  and  seconds  answering  to  any  given 
logarithmic  sine,  cosine,  tangent,  or  cotangent. 

Search  in  the  table,  and  in  the  proper  column,  until  the  num- 
ber be  found  ;  the  degrees  are  shown  either  at  the  top  or  bot- 
tom of  the  page,  and  the  minutes  in  the  side  columns,  either  at 
the  left  or  right.  But  if  the  number  cannot  be  exactly  found  in 
the  table,  take  the  degrees  and  minutes  answering  to  the  nearest 
less  logarithm,  the  logarithm  itself,  and  also  the  corresponding 
tabular  difference.     Subtract  the  logarithm  taken,  from  the 


PLANE  TRIGONOMETRY.  237 

given  logarithm,  annex  two  ciphers,  and  then  divide  the  re- 
mainder by  the  tabular  difference  :  the  quotient  is  seconds,  and 
is  to  be  connected  with  the  degrees  and  minutes  before  found  ; 
to  be  added  for  the  sine  and  tangent,  and  subtracted  for  the 
cosine  and  cotangent. 

Ex,  1.  To  find  the  arc  answering  to  the  sine       9.880054 
Sine  49°  20',  next  less  in  the  table,      ,     9.879963 


Tab.  DilF.  181)9100(50" 

Hence  the  arc  49°  20'  50"  corresponds  to.  the  given  sine 
9.880054. 

Ex.  2.  To  find  the  arc  corresponding  to  cotang.  10.008688. 

Cotang  44°  26',  next  less  in  the  table  10.008591 

Tab.  Diff:  421)9700(23" 

Hence,  44°26'— 23"=44°25'37"  is  the  arc  corresponding 
to  the  given  cotangent  10.008688. 


i 


PRINCIPLES  FOR  THE  SOLUTION  OF  RECTILINEAL  TRI- 
ANGLES. 

THEOREM  L 

In  every  right  angled  triangle,  radius  is  to  the  sine  of  either 
of  the  acute  angles,  as  the  hypothenuse  to  the  opposite  side  : 
and  radius  is  to  the  cosine  of  either  of  the  acute  angles,  as 
the  hypothenuse  to  the  adjacent  side. 

Let  ABC  be  the  proposed  tri- 
angle, right-angled  at  A  :  from 
the  point  C  as  a  centre,  with  a 
radius  CD  equal  to  the  radius  of 
the  tables,  describe  the  arc  DE, 
which  will  measure  tiie  angle  C  ; 
on  CD  let  fall  the  perpendicular 
EF,  which  will  be  the  sine  of  the 
angle  C,  and  CF  will  be  its  co- 
sine. The  triangles  CBA,  CEF,  are  similar,  and  give  the  pro- 
portion, 

CE  :  EF  :  :  CB  :  BA :  hence 

R  :  sin  C  :  :  BC  :  BA. 


238 


PLANE  TRIGONOMETRY 


But  we  also  have, 

CE  :  CF  : 
R  :  cos  C  : 


Cor. 


CB 
CB 


CA  :  hence 
CA. 


If  the  radius  R=l,  we  shall  have, 

AB  =  CB  sin  C,  and  CA=CB  cos  C. 

Hence,  in  every  right  angled  triangle,  the  perpendicular  is  equal 
to  the  hypothenuse  multiplied  by  the  sine  of  the  angle  at  the  base  ; 
and  the  base  is  equal  to  the  hypothenuse  multiplied  by  the  cosine 
of  the  angle  at  th,e  base  ;  the  radius  being  equal  to  unity. 


THEOREM  II. 


In  every  right  angled  triangle,  radius  is  to  the  tangent  of  ei- 
ther of  the  acute  angles,  as  the  side  adjacent  to  the  side  op- 
posite. 

Let  CAB  be  the  proposed  tri- 
angle. 

With  any  radius,  as  CD,  de- 
scribe the  arc  DE,  and  draw  the 
tangent  DG. 

From  the  similar  triangles 
CDG,  CAB,  we  shall  have, 

CD  :  DG  : :  CA  :  AB  :  hence, 
R  :  tang  C  : :  CA  :  AB. 

Cor.l.     If  the  radius  R=l, 

AB=CAtangC. 
Hence,  the  perpendicular  of  a  right  angled  triangle  is  equal  to 
the  base  multiplied  by  the  tangent  of  the  angle  at  the  base,  the 
radius  being  unity. 

Cor.  2.  Since  the  tangent  of  an  arc  is  equal  to  the  cotangent 
of  its  complement  (Art.  VI.),  the  cotangent  of  B  may  be  sub- 
stituted in  the  proportion  for  tang  C,  which  will  give 
R  :  cot  B  : :  CA  :  AB. 


THEOREM  III. 


In  every  rectilineal  triangle,  the  sines  of  the  angles  are  to  each 
other  as  the  opposite  sides. 


PLANE  TRIGONOMETRY. 


239 


Let  ABC  be  the  proposed  triangle  ;  AD 
the  pcrpcnciicula^  let  fall  from  the  vertex  A 
on  the  opposite  side  BC  :  there  may  be  two 
cases. 

First.    If  the  perpendicular  falls  within  -o 
the  triangle  ABC,  the  right-angled  triangles 
ABD,  ACD,  will  give, 

R  :  sin  B  : :  AB  :  AD. 
R  :  sin  C  : :  AC  :  AD. 

In  these  two  propositions,  the  extremes  are  equal ;  hence, 
sin  C  :  sin  E  : :  AB  :  AC. 

Secondly.  If  the  perpendicular  falls 
without  the  triangle  ABC,  the  right- 
angled  triangles  ABD,  ACD,  will  still 
give  the  proportions, 

R  :  sin  ABD  : :  AB  :  AD, 
R:sinC        ;:AC:AD; 

Trom  which  we  derive 

sin  C  :  sin  ABD 

But  the  angle  ABD  is  the  supplement  of  ABC,  or  B ;  hence 
sin  ABD = sin  B ;  hence  we  still  have 

sinC:sinB;:AB:AC. 


AB :  AC. 


THEOREM  IV. 


In  every  rectilineal  triangle,  the  cosine  of  either  of  the  angles  is 
equal  to  radius  multiplied  by  the  sum  of  the  squares  of  the  sides 
adjacent  to  the  angle,  minus  the  square  of  the  side  opposite, 
divided  by  twice  the  rectangle  of  the  adjacent  sides. 

Let  ABC  be  a  triangle :  then  will 

AB^+BC2— AC2 


cos  B=R- 


2ABxBC. 


First.  If  the  perpendicular  falls  within 

,'the  triangle,  we  shall  have  AC^^rAB^H- 

BC2— 2BCxBD(Book  IV.  Prop.  XII.);  B 

.          T3^,    AB^+BC^— AC2    ^  ,  .     ^     .  ^         ,  ,     .      , 
I  hence  BD=: —.^^^ .   But  m  the  right-angled  triangle 


jABD,  we  have 


2BC 


R :  cos  B  : :  AB  :  BD ; 


^40  PLANE  TRIGONOMETRY. 

hence,  cos  B—    ^p    ,  or  by  substituting  the  value  of  BD, 
AB2+BC2_AC2 


cos  B— R  X 


2AB  X  BC 


Secondly.    If  the  perpendicular  falls 
without  the  triangle,  we    shall    have 
AC2z=:AB2+BCH2BCxBD;    hence 
p^    AC^—AB^— BC2 

^^== 2BC 

But  in  the  right-angled  triangle  BAD,  

RxBD  D     B            C 

we  still  have  cos  ABD=:— -^g-  ;  and  the  angle  ABD  being 

supplemental  to  ABC,  or  B,  we  have 

cosB=-cosABD=-?-f|^. 

AB 

hence  by  substituting  the  value  of  BD,  we  shall  again  have 

P     „     AB2+BC2— AC2 

""^^=^><— 2AB^BG-- 

Scholium.   Let  A,  B,  C,  be  the  three  angles  of  any  triangle  ; 
a,  h,  c,  the  sides  respectively  opposite  them :  by  the  theorem, 

we  shall  have  cos  B=R  x  — ^ .     And  the  same  principle, 

when  applied  to  each  of  the  other  two  angles,  will,  in  like  man- 

ner  give  cos  A=R  x  — kt ,  anc  cos  C=R  x  — tt-t — . 

*=  26c  2ab 

Eithei-  of  these  formulas  may  readily  be  reduced  to  one  in  which 

the  computation  can  be  made  by  logarithms. 

Recurring  to  the  formula  R'-^ — R  cos  A=2sin^  ^A  (Art. 

XXIII.),  or  2sin^^A=R~ — RcosA,  and  substituting  for  cosA, 

we  shall  have 

2sin^^ArrR^-^R^x     \^^ 

Wx2bc--n\h^+c^-^')  a^^b''-^c^+2bc 

2bc  — R  X  2^^ 

^  R.xf!:^)Wx  (^^=4M£l±).  ,  Hence 
2oc  2bc 

^  46c  ^ 

For  the  sake  of  brevity,  put 

J  (a  +  /;  +  c)  =/?,  or  a + 6 + c  =2p ;  we  have  a + h — c=2p — 2c, 
a-\-c — 6=2p — 2h ;  hence 


a.nJA=RV(i^=^^)). 


PLANE  TRIGONOMETRY.      '''"'^'^241 


THEOREM  V. 

In  every  rectilineal  triangle,  the  sum  of  two  sides  is  to  their  diffe- 
rence as  the  tangent  of  half  the  sum  of  the  angles  opposite  those 
sides,  to  the  tangent  of  half  their  difference. 

For.  AB  :  BC  :  :  sin  C  :  sin  A  (Theo- 
rem III.).  Hence,  AB  +  BC  :  AB— BC 
; :  sin  C  -t-sin  A  :  sin  C — sin  A.         But 

.  ^      .    .      .   ^      .    .  C  +  A 

sinC  +  sinA:  sinC — sin  A  : :  tang—— —  : 

tang  — ^r—  (Art.  XXIV.)  ;  hence, 

AB+BC  :  AB— BC  :  :  tang  ^±^  :  tang  "IZH^,   which  is 

fi  <« 

the  property  we  had  to  demonstrate. 

With  the  aid  of  these  five  theorems  we  can  solve  all  the 

cases  of  rectilineal  trigonometry. 

Scholium.  The  required  part  should  always  he  found  from 
the  given  parts ;  so  that  if  an  error  is  made  in  any  part  of  the 
work,  it  may  not  affect  the  correctness  of  that  which  follows. 


SOLUTION  OF  RECTILINEAL  TRIANGLES  BY  MEANS  OF 
LOGARITHMS. 

It  has  already  been  remarked,  that  in  order  to  abridge  the 
calculations  which  are  necessary  to  find  the  unknown  parts  of 
a  triangle,  we  use  the  logarithms  of  the  parts  instead  of  the 
parts  themselves. 

Since /the  addition  of  logarithms  answers  to  the  mqltiphca- 
tion  of  tneir  corresponding  numbers,  and  their  subtraction  to 
the  division  of  their  number^;  it  follows,  that  the  logarithm  of 
the  fourth  term  of  a  proportion  will  be  equal  to  the  sum  of 
the  logarithms  of  the  second  and  third  terms,  diminished  by 
the  logarithm  of  the  first  term. 

Instead,  however,  of  subtracting  the  logarithm  of  the  first 
term  from  the  sum  of  the  logarithms  of  the  second  and  third 
terms,  it  is  more  convenient  to  use  the  arithmetical  complement 
of  the  first  term. 

/The  arithmetical  complement  of  a  logarithm  is  the  number 
which  remains  after  subtracting  the  logarithm  from  10. J  Thus 
10 — 9.274687  =  0.725313:  hence,  0.725313  is  the  arithmetical 
complement  of  9.274687.      X 


242  PLANE  TRIGONOMLrUY. 

It  is  now  to  be  shown  that,  the  difference  hcfvhen  rvo  V/'if 
r'lthms  is  truly  founds  hij  adding  to  the  first  logarxihm  ih-°  ^'rith- 
meticdl  complement  of  the  logarithm  to  he  subtracledi  end  dimin- 
ishing their  sum  by  10. 

Let  a  =z  the  first  logarithm. 

b  =  the  logarithm  to  be  subtracted. 

c  =  10 — b=xhe  arithmetical  complement  of  6, 

Now,  the  difference  between  the  two  logarithms  will  be 
expressed  by  a — b.  But  from  the  equation  c=10 — b,  we  have 
c — 10= — b  :  hence  if  we  substitute  for  ^-b  its  value,  we  shall 
have 

a — b=a-{-c — 10, 

which  agrees  with  the  enunciation. 

/  When  we  wish  the  arithmetical  complement  of  a  logarithm, 
/we  may  write  it  directly  from  the  tables,  by  subtracting  the 
/  left  hand  figure  from  9,  then  proceeding  to  the  right,  subtract 
f  each  figure  from  9,  till  we  reach  the  last  significant  figure,  which 
\    must  be  taken  from  10 :  this  will  be  the  same  as  taking  the 

logarithm  from  10. 

Ex.  From  3.274107  take  2.104729. 

Common  method.  By  ar.-comp. 

3.274107  3.274107 

2.104729  ar.-comp.  7.895271 


Diff.    1.169378  sum    1.169378  after    re- 

jecting  the  10. 

We  therefore  have,  for  all  the  proportions  of  trigonometry, 
tlie  following 

RULE. 

Add  together  the  arithmetical  complement  of  the  logarithm  of  the 
the  first  term,  the  logarithm  of  the  second  term,  and  the  loga- 
rithm of  the  third  term,  and  their  sum  after  rejecting  10,  will 
he  the  logarithm  of  the  fourth  term..  And  if  any  expression 
occurs  in  which  the  arithmeticcd  com,plement  is  twice  used,  20 
must  be  rejected  fivm  the  sum. 


i 


c 


PLANE  TRIGONOMETRY.  243 


SOLUTION  OF  RIGHT  ANGLED  TRIANGLES. 

Let  A  be  the  right  angle  of  the  proposed 
right  angled  triangle,  B  and  C  the  other  two 
angles ;  let  a  be  the  hypothenuse,  b  the  side 
opposite  the  angle  B,  c  the  side  opposite  the 
angle  C.  Here  we  must  consider  that  the  ^ 
two  angles  C  and  B  are  complements  of  each  other  ;  and  that 
consequently,  according  to  the  different  cases,  we  are  entitled 
to  assume  sin  C^=cos  B,  sin  B=cos  C,  and  likewise  tang  B= 
coi  C,  tang  C= cot  B.  This  being  fixed,  the  unknown  parts 
of  a  right  angled  triangle  may  be  found  by  the  first  two  theo- 
rems; or  if  two  of  the  sides  are  given,  by  means  of  the  pro- 
perty, that  the  square  of  the  hypothenuse  is  equal  to  the  sum 
of  the  squares  of  the  other  two  sides. 

EXAMPLES.  '^ 

Ex.  1.  In  the  right  angled  triangle  BCA,  there  are  given  t!:e 
hypothenuse  (2=250,  and  the  side  6=240 ;  required  the  other 
parts. 

R  :  sin  B  :  :  a  :  6  (Theorem  L). 
or,  «  :  6  :  :  R  :  sin  B. 

•  When  logarithms  are  used,  it  is  most  convenient  to  write  tlie 
proportion  thus. 

As  hyp.  a      -     250 '     -      ar.-comp.     log.     -     7.602060 

To  side/;      -     240 2.380211 

So  is      R 10.000000 

To  sin  B      -      73°  44' 23"  (after  rejecting  10)    9.982271 

But  the  angle  C=90°— B=90^— 73°  44'23"=16°  15'  37" 
or,  C  might  be  found  by  the  proportion, 

As  hyp.  a  -     250     -     ar.-comp.          log.     -  7.002060 

To  side  /?     -     240 -     -  2.380211 

So  is      R -     -  lO.OOOOGO 

To  cos  C     -       16°  15' 37" 9.982271 

To  find  the  side  c,  we  say, 

As  R        -         -         ar.  comp.  log.  -  0.000000 

To  tang.  C   16°  15'  37"      -  -         -  -  9.464889 

So  is  side  6  240         ;         -  -  -  2.380211 

To  side      c  70.0003           -  -  -  1.845100 


214  PLANE  TRIGONOMETRY. 

Or  the  side  c  might  be  found  from  the  equation 

For,  c^=a^—h''={a-\-h)x{a—b): 

hence,  2log.  c=log.  (a+t)  +  log.  (a — ^,  or 

log.  c=Jlog.  {a-\~h)  ^l\og.  (a-^b) 
a4-5=250  +  240=490       log.         2.690196 
a— &=250— 240=10     -         -         1.000000 

2)  3.690196 
Log.  c      70       ---..-         -         1.845098 

Ex.  2.  In  the  right  angled  triangle  BCA,  there  are  given, 
side 6=384 yards,  and  the  angle  B=53°  8' :  required  the  other 
parts. 

To  find  the  third  side  c. 
R  :  tang  B  :  :  c  :  b  (Theorem  II.) 
or,  tangB  :  R  :  :  h  :  c.     Hence, 

As  tang     B  53°  8'             ar.-comp.  log.  9.875010 

is  to          R              ...         -  -  10.000000 

So  is  side  b  384       -         -         -         -  .  2.584331 

To  side           c  287.965         -         -         -  -  "2^459341 

Note.  When  the  logarithm  whose  arithmetical  complement 
is  to  be  used,  exceeds  10,  take  the  arithmetical  complement 
with  reference  to  20  and  reject  20  from  the  sum. 

To  find  the  hypothenuse  a. 
R  :  sin  B  :  :  a  :  6    (Theorem  I.).     Hence, 

As  sin  B  53°  8'         ar.  comp.  log.  0.096892 

Is  to  R              -         -         -  -  -  10.000000 

So  is  side  6.384         -         -  -  -  2.584331 

To  hyp.     a  479.98             -  -  -  2.681223 

Ex.  3.  In  the  right  angled  triangle  BAG,  there  are  given, 
side  c=195,  angle  B=47°  55', 
required  the  other  parts. 

Ans.  Angle  G=42°  05',  a==290.953^  6=215.937. 


SOLUTION  OF  RECTILINEAL  TRIANGLES  IN  GENERAL. 

Let  A,  B,  G  be  the  three  angles  of  a  proposed  rectilineal  tri- 
angle ;  ff,  6,  c,  the  sides  which  are  respectively  opposite  them  ; 
the  different  problems  which  may  occur  in  determining  three  of 
these  quantities  by  means  of  the  other  three,  will  all  be  redu- 
cible to  the  four  following  cases. 


TLANE  TRIGONOMETRY.  245 


CASE  I. 

Given  a  side  and  two  angles  of  a  triangle,  to  find  the  remaining 

parts. 

First,  subtract  the  sum  of  the  two  angles  from  two  right  an- 
gles, the  remainder  will  be  the  third  angle.  The  remaining 
sides  can  then  be  found,  by  Theorem  111. 

I.  In  the  triangle  ABC,  there  are  given  the  angle  A =58°  07', 
the  angle  B  =  22°  37',  and  the  side  07=  408  yards:  required  ih^ 
remaining  angle  and  the  two  other  sides. 

To  the  angle  A =58°  OT 

Add  the  angle  B       -        ...        -         =22°  37' 

Their  sum  -         -         -         -        -         -         =80°  44' 

taken  from  180°  leaves  the  angle  C  -         =99°  10'. 

This  angle  being  greater  than  90°  its  sine  is  found  by  taking 
that  of  its  supplement  80°  44'. 


To  find  the  side  a. 

As  sine  C 
Is  to  sine  A 
So  is  side  c 

99°  16' 

58°  07' 
408     - 

ar.-comp. 

log. 

0.005705 
9.928972 
2.610660 

So  side  a 

351.024 

- 

- 

2.545337 

To  find  the  side  h. 

As  sine  C 
Is  to  sine  B 
So  is  side  c 
To  side  h 

99°  16' 
22°  37' 
408     - 
158.976 

ar.-comp. 

log. 

0.005705 
9.584968 
2.610660 
2.201333 

2.  In  a  triangle  ABC,  there  are  given  the  angle  A  =  38°  25' 
B  =  57°  42',  and  the  side  c=400  :  required  the  remaining 
parts. 

Ans,  Angle  C=83°  53',  side  a=249.974,  side  &=340.04. 

CASE  II.         Y 

Given  two  sides  of  a  triangle,  and  an  angle  opposite  one  of  them, 
to  find  the  third  side  and  the  two  remaining  angles, 

X* 


246 


PLANE  TRIGONOMETRY. 


1 .  In  the  triangle  ABC,  there 
are  given  side  AC  =  21 6,  BC=: 
117,  and  the  angle  A =22°  37', 
to  find  the  remaining  parts. 

Describe  the  triangles  ACB, 
ACB',  as  in  Prob.  XI.  Book  III. 

Then  find  the  angle  B  by- 
Theorem  III. 

As  sideB'C  or  BC  117 
Isioside  AC  216 

So  is  sine  A  22°  37' 

To  sine  B'  45°  13'  55"  or  ABC  134°  46'  05" 

Add  to  each  A    22°  37' 00"  22°  37'  00'^ 

Take  their  sum    67"  50  5'5 "  '  157°  23' 05" 

From  180°  00' 00"  180°  00' 00" 

Rem.     ACB'     112°  0905"     ACB=22°  36'  55" 


ar.-comp.     log. 


7.931814 
2.334454 
9.584968 
9.851236 


To  find  the  side  AB  or  AB'. 


As  sine  A  22°  37'         ar.-comp. 

Is  to  sine  ACB'  112°  09' 05"      - 
So  is  side  B'C    117        - 

To  side  AB'     281.785 


log. 


0.415032 
9.966700 
2.068186 
2.449918 


The  ambiguity  in  this,  and  similar  examples,  arises  in  con- 
sequence of  the  first  proportion  being  true  for  both  the  trian- 
gles ACB,  ACB'.  As  long  as  the  two  triangles  exist,  the  am- 
biguity will  continue.  But  if  the  side  CB,  opposite  the  given 
angle,  be  greater  than  AC,  the  arc  BB'  will  cut  the  line  ABB', 
on  the  same  side  of  the  point  A,  but  in  one  point,  and  then 
there  will  be  but  one  triangle  answering  the  conditions. 

If  the  side  CB  be  equal  to  the  perpendicular  Crf,  the  arc 
BB'  wil,l  be  tangent  to  ABB',  and  in  this  case  also,  there  will 
be  but  one  triangle.  When  CB  is  less  than  the  perpendicular 
Cd,  the  arc  BB'  will  not  intersect  the  base  ABB',  and  in  that 
case  there  will  be  no  triangle,  or  the  conditions  are  impossible. 

2.  Given  two  sides  of  a  triangle  50  and  40  respectively,  and 
the  angle  opposite  the  latter  equal  to  32° :  required  the  remain- 
ing parts  of  the  triangle. 

-4715.  If  the  angle  opposite  the  side  50  be  acute,  it  is  equal 
to  41°  28'  59",  the  third  angle  is  then  equal  to  106°  31'  01",  and 
the  third  side  to  72.368.  If  the  angle  opposite  the  side  50  be 
obtuse,  it  is  equal  to  138°  31'  01",  the  third  angle  to  9°  28'  59', 
and  the  remaining  side  to  12.436. 


PLANE  TRIGONOMETRY.  247 


CASE  III. 

Given  two  sides  of  a  triangle^  with  their  included  angle,  to  find 
the  third  side  and  the  two  remaining  angles. 

Let  ABC  be  a  triangle,  B  the  given 
angle,  and  c  and  a  the  given  sides. 

Knowing  the  angle  B,  we  shall  like- 
wise know  the  sum  of  the  other  two  an- 
gles C  +  A=180°— B,  and  their  half  sum       

J  (C  +  A)=90— JB.     We    shall    next    A!  b  *C 

compute  the  half  difference  of  these  two  angles  by  the  propor- 
tion (Theorem  V.), 

c  +  a  :  c — a  : :  tang  J  (C  +  A)  or  cot  J  B  :  tang  J  (C — A,) 

in  which  we  consider  c>a  and  consequently  C>A.  Having 
found  the  half  difference,  by  adding  it  to  the  half  sum 
^  (C  +  A),  we  shall  have  the  greater  angle  C  ;  and  by  subtract- 
ing it  from  the  half-sum,  we  shall  have  the  smaller  angle  A. 
For,  C  and  A  being  any  two  quantities,  we  have  always, 

Crz:i(C  +  A)+^(C— A) 
A::z:i(C+A)-J(C-A). 

Knowing  the  angles  C  and  A  to  find  the  third  side  h,  we  liave 
the  proportion. 

sin  A  :  sin  B  : :  a  :  & 

Ex.  L  In  the  triangle  ABC,  let  a =450,  c=:540,  and  the  in- 
cluded angle  B=z:  80"^ :  required  the  remaining  parts. 

c  +  a— 990,c— a  =  90,  180"— B  =  100°=C  +  A. 

Asc  +  a           990             ar.-comp.         log.  7.004365 

Is  toe— a         90 1.954243 

So  is  tang  J  (C  +  A)  50°           -.         -         -  10.076187 

To  tangi  (C— A)  6°  11'             -         -         -  "9.0;i4795 

Hence,  50°  +  6°  ll'  =  56°  ll'^C;  and  50^—6°  ir=43''  49' 
=A. 

To  find  the  third  side  b. 

As  sine  A       43"  49'         ar.-comp.         log.         0.159672 

Is  to  sine  ^80° 9.993351 

So  is  side  a     450      -  -         .         -         2.653213 

To  side  b        640.082        -         -         -         .         2r8"06236 
Ex.  2.     Given  two  sides  of  a  plane  triangle,  1686  and  960, 
and  their  included  angle  128°  04':  required  the  other  parts. 

Ans.  Angles,  33°  34'  39 ",  18°  21'  21",  side  2400. 


248  PLANE  TRIGONOMETRY. 

CASE  IV. 
Given  the  three  sides  of  a  triangle,  to  find  the  angles. 
We  have  from  Theorem  IV.  the  formula, 

.in -J  A=R^(li^(Z^     in  which 
p  represents  the  half  sum  of  the  three  sides.     Hence, 
sn.^A=w(P-'Y/-'),    or 

2  log.  sin  JA=2  log.  R+log.  (p — Z>)-l-log.  (p — c) — log.  c — 
log.  b, 

Ex.  1.     In  a  triangle  ABC,  let  b=40,  c=34,  and  a=25: 
required  the  angles. 

40  +  34  +  25  ,     ^ 

Here  jo= =^49.5,  p — 0=9.5,  and  p — c=15.5. 

2  Log.  R         -  -         -        -         -         -  20.000000 

log.  ip—h)     9.5  -         -         -         -         -  0.977724 

log.  (p-^c)  15.5  -         -         -         -         -  1.190332 

— log.  c           34             ar.-comp.       -         -  8.468521 

— log.  h           40             ar.-comp.       -         -  8.S97940 

2  log.  sin  J  A 19.034517 

log.  sin  J  A  19°  12'  39''  -         -         -  9-517258 

Angle  A=38°  25'  18". 

In  a  similar  manner  we  find  the  angle  B=83°  53'  18"  and 
the  angle  C=57°  41'  24". 

Ex.  2.  What  are  the  angles  of  a  plane  triangle  whose  sides 
are,  a=60,  ^=50,  and  c=40? 

Ans.  41°  24'  34",  55°  46'  16"  and  82°  49'  10". 


APPLICATIONS. 

Suppose  the  height  of  a  building  AB  were  tequired,  the 
foot  of  it  being  accessible. 


PLANE  TRIGONOMETRY. 


249 


On  the  ground  which  we 
suppose  to  be  horizontal  or  very 
nearly  so,  measure  a  base  AD, 
neither  very  great  nor  very 
small  in  comparison  with  the 
altitude  AB ;  then  at  D  place 
the  foot  of  the  circle,  or  what- 
ever be  the  instrument,  with 
which  we  are  to  measure  the 

angle  BCE  formed  by  the  hori-  

zontal  line  CE  parallel  to  AD,       A  I> 

and  by  the  visual  ray  direct  it  to  the  summit  of  the  building. 
Suppose  we  find  AD  or  CErr67.84  yards,  and  the  angle 
BCE^=41"  04' :  in  order  to  find  BE,  we  shall  have  to  solve 
the  right  angled  triangle  BCE,  in  which  the  angle  C  and  the 
adjacent  side  CE  are  known. 


To  find  the  side  EB. 


AsR 


ar.-comp. 


0.000000 


Is  to  tang.  C  41°  04' ".     .     .     9.940183 

So  is  EC  67.84 1.831486 


ToEB 


59.111 1.771669 


Hence,  EB=59.111  yards.  To  EB  add  the  height  of  the 
instrument,  which  we  will  suppose  to  be  1.12  yards,  we  shall 
then  have  the  required  height  AB=60.231  yards. 

If,  in  the  same  triangle  BCE  it  were  required  to  find  the 
hypothenuse,  form  the  proportion 
'    As  cos  C  41°  04'        ar.-comp.        -     -       log.     0.122660 

Is  to  R :     -     -     -     10.000000 

So  is  CE   67.84 1.831486 

ToCB       89.98     ---..-...       1.954146 


Note,  If  only  the  summit  B  of  the  building  or  place  whose 
height  is  required  were  visible,  we  should  determine  the  dis- 
tance CE  by  the  method  shown  in  the  following  example ; 
this  distance  and  the  given  angle  BCE  are  sufficient  for  solv- 
ing the  right  angled  triangle  BCE,  whose  side,  increased  by 
the  height  of  the  instrument,  will  be  the  height  required. 


250 


PLANE  TRIGONOMETRY. 


2.  To  find  upon  the  ground 
the  distance  of  the  point  A 
from  an  inaccessible  object 
B,  we  must  measure  a  base 
AD,  and  the  two  adjacent 
angle?  BAD,  ADB.  Sup- 
pose  we  have  found  AD= 
588.45  yards,  BAD  =  103° 
55'  55",  and  BD A  =  36°  04'; 
we  shall  thence  get  the  third 
angle  ABD=40°  05",  and  to 
obtain  AB,  we  shall  form  the 
proportion 

As  sine  ABD  40°  05" 
Is  to  sin  BDA  36°  04' 
So  is  AD  588.45 

To  AB  -    -    538.943 


ar.-comp. 


log. 


-  0.191920 

-  9.709013 

-  2.769710 


2.731543 


If  for  another  inaccessible  object  C,  we  have  found  the  an- 
gles CAD=35°  15',  ADC  =  119°  32',  we  shall  in  like  manner 
find  the  distance  AC  =  1201.744  yards. 

3.  To  find  the  distance  between  two  inaccessible  objects  C 
and  C,  we  determine  AB  and  AC  as  in  the  last  example  ;  we 
shall,  at  the  same  time,  have  the  included  angle  BAC  =  BAD — 
DAC.  Suppose  AB  has  been  found  equal  to  538.818  yards, 
AC  =  1201.744  yards,  and  the  angle  BAC  =  08°  40'  55";  to 
get  BC,  we  must  resolve  the  triangle  BAC,  in  which  are  known 
two  sides  and  the  included  angle. 

AsAC  +  AB     1740.562     ar.-comp.        log.-     6.759311 

Is  to  AC— AB   662.926 2.821465 

P  -4-  C 

So  is  tang.— ^ .  55°  39' 32"     -     -     -     -     -10.165449 


To  tang. 


B- 


2 
-C 


29°  08'  19" 9.74u225 


Hence    -    -    -    - 
But  wc  have    -    - 


B— C 


2 
B  +  C 


:29°  OS'  19 


=  55°  39'  32' 


Hence    -    -    -*-    -    -  B     =84°  47' 51" 
and         C     =26°  31'  13" 


PLANE  TRIGONOMETRY.  25J 

Now,  to  find  the  distance  BC  make  the  proportion, 

As  sine  B  84^  47'  51"     ar.-comp.     -     log.     -  0.001793 

Is  to  sine  A  08'"  40'  55 ' 9.909218 

So  is  AC  1201.744        -     - 8.079811 

To  BC  1124.145      -     -     -. 3.050822     ^ 

4.  Wanting  to  know  the  distance  between  two  inaccessible 
objects  which  lie  in  a  direct  line  from  the  bottom  of  a  tower 
of  120  feet  in  height,  the  angles  of  depression  are  measured, 
and  found  to  be,  of  the  nearest,  57° ;  of  the  most  remote, 
25^  30' ;  required  the  distance  between  them. 

Ans.  173.G5G  feet. 

5.  In  order  to  find  the  distance  between  two  trees,  A  and 
B,  which  could  not  be  directly  measured  because  of  a  pool 
which  occupied  the  intermediate  space,  the  distance  of  a  third 
poi'.it  C  from  each,  was  measured,  viz.  CA=588  feet  and  CB 
=072  feet,  and  also  the  contained  angle  ACB=55°  40':  requi- 
red the  distance  AB. 

Ans.  592.907  feet. 

6.  Being  on  a  horizontal  plane,  and  wanting  to  ascertain 
the  height  of  a  tower,  standing  on  the  top  of  an  inaccessibje 
hill,  there  were  measured,  the  angle  of  elevation  of  the  top  of 
the  hill  40^  and  of  the  top  of  the  tower  51°  :  then  measuring 
in  a  direct  line  180  feet  farther  from  the  hill,  the  angle  of  ele- 
vation of  the  top  of  the  tower  was  33°  45' :  required  the  height 
of  *he  tower. 

Ans.  83.9983  feet. 

7.  Wanting  to  know  the  horizontal  distance  between  two 
inaccessible  objects  A  and  B,  and  not  finding  any  station  from 
which  both  of  them  could  be  seen,  two  points  C  and  D,  were 
chosen,  at  a  distance  from  each  other  equal  to  200  yards,  from 
the  former  of  which  A  could  be  seen,  and  from  the  latter  B, 
and  at  each  of  the  points  C  and  D  a  staff  was  set  up.  From 
C  a  distance  CF  was  measured, not  in  the  direction  DC,  equal 
to  200  yards,  and  from  D,  a  distance  DE  equal  to  200  yards, 
and  the  following  angles  were  taken,  viz.  AFC=83°  ACF= 
54°  31',  ACD=:53°  30 ,  BDC=156°  25',  BDE=54°  30',  and 
BED =88°  30' :  required  the  distance  AB. 

Ans.  345.46  yards. 

8.  From  a  station  P  there  can  be  seen  three  objects.  A,  B 
and  C,  whose  distances  from  each  other  are  known,  viz.  AB= 
800,  AC=:600,  and  BC=400  yards.  There  are  also  measured 
the  horizontal  angles,  A?C  =  33°  45',  BPC=22°  30'.  It  is  re- 
quired, from  these  data,  to  determine  the  three  distances  PA» 
PC  and  PB. 

Ans.  PA=710.193,  PC=:1042.522,  PB=934.291  yards. 


> 


252  SPHERICxVL  TRIGONOMETRY. 


SPHERICAL  TRIGONOMETRY. 

I.  It  has  already  been  shown  that  a  spherical  triangle  is 
formed  by  the  arcs  of  three  great  circles  intersecting  each  other 
on  the  surface  of  a  sphere,  (Book  IX.  Def.  1).  Hence,  every 
spherical  triangle  has  six  parts :  the  sides  and  three  angles. 

Spherical  Trigonometry  explains  the  methods  of  determin- 
ing, by  calculation,  the  unknown  sides  and  angles  of  a  spheri- 
cal triangle  wh-en  any  three  of  the  six  parts  are  given. 

II.  Any  two  parts  of  a  spherical  triangle  are  said  to  be  of 
the  same  species  when  they  are  both  less  or  both  greater  than 
90*^ ;  and  they  are  of  different  species  when  one  is  less  and  the 
other  greater  than  90°. 

III.  Let  ABC  be  a  spherical  JV 
triangle,  and  O  the  centre  of  the  /l\^\^ 
sphere.  Let  the  sides  of  the  tri-  /  |\    ^"^^ 
angle  be  designated  by  letters  /    ||  \          ^^^^^^ 
cori-esponding  to  their  opposite  /     lV\  \           ^  ^T^^^ 

angles :  that  is,  the  side  opposite  B  [ — rtrr^ ¥ — H"t^O 

the  angle  A  by  a,  the  side  oppo-  \i-i/\    \       4i      ^y^ 

site  B  by  i,  and  the  side  opposite  \  '<:t"- """"j'ii      y^ 

C  by  c.     Then  the  angle  COB  \j\'-<\'ly^ 

will  be  represented  by  a,  the  an-  \\  y^ 

gle  CO  A  by  h  and  the  angle  ^q 

BOA  by  c.     The  angles  of  the 

spherical  triangle  will  be  equal  to  the  angles  included  between 

the  planes  which  determin.3  its  sides  (Book  IX.  Prop.  VI.). 

From  any  point  A,  of  the  edge  OA,  draw  AD  perpendicular 
to  the  plane  COB.  From  D  draw  DH  perpendicular  to  OB, 
and  DK  perpendicular  to  OC ;  and  draw  AH  and  AK :  the 
last  lines  will  be  respectively  perpendicular  to  OB  and  OC, 
(Book  VI.  Prop.  VI.) 

The  angle  DH  A  will  be  equal  to  the  angle  B  of  the  spheri- 
cal triangle,  and  the  angle  DKA  to  the  angle  C. 

The  two  right  angled  triangles  OICA,  ADK,  will  give  the 
proportions 

R  :  sin  AOK  : :  OA  :  AK,  or,  R  x  AKrr  OA  sin  b. 
R  :  sin  AKD  : :  AK  :  AD,  or,  R  x  AD=AK  sin  C. 

Hence,  R^  x  AD=AO  sin  h  sin  C,  by  substituting  for  AK  its 
value  taken  from  the  first  equation. 


SPHERICAL  TRIGONOMETRY.  253 

In  like  manner  the  triangles  AHO,  ADH,  right  angled  at 
H  and  D,  give 

R  :  sin  c  : :  AO  :  AH,  or  R  x  AH= AO  sin  c 
R  :  sin  B  : :  AH  :  AD,  or  R  x  ADr=:AH  sin  B. 
Hence,         R^  x  AD= AO  sin  c  sin  B. 

Equating  this  with  the  value  of  R^  x  AD,  before  found,  and  di- 
viding by  AO,  we  have 

.    ^      •         .    T.       sin  C     sin  c 
sm  b  sm  C=:sin  c  sm  B,  or  - — ^=-^ — r     ( 1 ) 
'      sin  B     sin  6     ^    ' 

or,        sin  B  :  sin  C  : :  sin  6  :  sin  c  that  is, 

The  sines  of  the  angles  of  a  spherical  triangle  are  to  each 
other  as  t/ie  sines  of  their  opposite  sides. 

IV.  From  K  draw  KE  perpendicular  to  OB,  and  from  D  draw 
DF  parallel  to  OB.  Then  will  the  angle  DKF=:COB=a, 
since  each  is  the  complement  of  the  angle  EKO. 

In  the  right  angled  triangle  OAH,  we  have 

R  :  cos  c  : :  OA  :  OH  ;  hence         * 
AO  cos  c=RxOH=RxOE+R.DF. 

In  the  right-angled  triangle  OKE 

R  :  cos  a  : :  OK  :  OE,  or  Rx OE=OK  cos  a. 
But  in  the  right  angled  triangle  OKA 

R  :  cos  6  : :  OA  :  OK,  or,  Rx  OK=OA  cos  b, 

vy  TT  TJ       /^T^       r^  A    COS  «  cos  b 

^  Hence     RxOE=OA. ^ 

In  the  right-angled  triangle  KFD 

R  :  sin  «  :  KD  :  DF,  or  R  X  DF=KD  sin  a. 

But  in  the  right  angled  triangles  OAK,  ADK,  we  have 

R  :  sin  6  : :  OA  :  AK,  or  Rx  AK=OA  sin  b 

R  :  cos  K  :  AK  :  KD,  or  R  x  KD- AK  cos  C 

,                  T^Tx     ^A  sin  b  cos  C 
hence        KD  = ^ ,  and 

OA  sin  a  sin  &  cos  C 
K  X  Dr  = j^2 :  therefore  * 

OA  cos  «  cos  6    AO  sin  a  sin  b  cos  C 
OA  cos  c= ^ + =^2 1  ^ 

R^  cos  c— R  cos  a  cos  &+sin  a  sin  b  cos  C. 
Y 


254  SPHERICAL  TRIGONOMETRY. 

Similar  equations  may  be  deduced  for  each  of  the  other 
sides.     Hence,  generally, 


R^  cos  a=R  cos  b  cos  c+sin  b  sin  c  cos  A. 
R2  cos  b=p,  cos  a  cos  c  +  sin  a  sin  c  cos  B. 
R^  cos  c==R  cos  b  cos  a  +  sin  b  sin  a  cos  C. 


That  is,  radius  square  into  the  cosine  of  either  side  of  a  spheri- 
cal triangle  is  equal  to  radius  into  the  rectangle  of  the  cosines  of 
the  two  other  sides  plus  the  rectangle  of  the  sines  of  those  sides 
into  the  cosine  of  their  included  angle. 

V.  Each  of  the  formulas  designated  (2)  involves  the  three 
sides  of  the  triangle  together  with  one  of  the  angles.  These 
formulas  are  used  to  determine  the  angles  when  the  three  sides 
are  known.  It  is  necessary,  however,  to  put  them  under  an- 
other form  to  adapt  them  to  logarithmic  computation. 

Taking  the  first  equation,  we  have 

R^  cos  a — R  cos  b  cos  c 


cos  A=- 


sin  b  sin  c 


Adding  R  to  each  member,  we  have 

R2  cos  a-f  R  sin  b  sin  c — R  cos  6  cos  c 


R+cos  A: 


sin  b  sin  c 


But,    R  +  cos  A^p^^       (Art.  XXIII.),  and 

■R  sin  b  sin  c— R  cos  b  cos  c= — R^  cos  (6  +  c)  (Art.  XIX.) ; 
-            2  cos^^A     R2  (cos  a — cos(6  +  c)) 
hence, g—  = sin  b  sin  c = 

.  ^^  sin|  (a  +  Z>  +  c)  sin^  (b^c-a)  ^^^^^ 

sm  0  sm  c  ^ 

Putting  s = a + 5  +  c,    we  shall  have 

i5=i(a+6+c)  and  is— a=J  (&+c— a)  :  hence 

cos  i  K=Yisy^^SMWSI^\ 

^  ▼  cm   h  cin   r 


sin  b  sin  c 

.  ^     T»       /^in  i 
»cos 


sm  a  sm  c 


)■  (3-) 


cos  J  C=R^— ^-^5^,-55^6 


SPHERICAL  TRIGONOMETRY. 


255 


Had  we  subtracted  each  member  of  the  first  equation  from 
R,  instead  of  adding,  we  should,  by  making  similar  reductions, 
have  found 


sm 


sm 


sm 


/sin  ^(a  +  b — c)  sin  -|-  (a  +  c — b) ^ 


sin  b  sin  c 


/sin  l(a  +  b — c)  sin  ^  (b  +  c — a) 
2^I3=Kv  sin  a  sine 


n4.) 


sin  a  sin  6 

Putting  s=a+6  +  c,we  shall  have 

^s — a=^{b+c — a),  ^ — 6=J  {a+c — 6),  and  Js — c=^{a+b — c) 

hence, 


sin  1A=R.  Anft^-^)^ina»-ft) 
sin  6  sin  c 

sin  ^R^R^/sm  (^s— c)  sin  ftg=^) 


M5.) 


sm  a  sm  c 


sin  XCr=R.  /sin  (1^— ^)  sin  (^g— a) 
^  sin  a  sin  6  -^ 

VI.    We  may  deduce  the  value  of  the  side  of  a  triangle  in 

terms   of   the   three  angles    by  applying .  equations   (4.),  to 

'    the  polar  triangle.     Thus,  if  a';  ft',  c',  A',  B',  CVrepresent  the 

I  sides  and  angles  of  the  polar  triangle,  we  shall  have 
A==180°— ff',  B  =  180°— &',  C=180°— c' ; 
a= 180°— A',  Z>= 180°— B',  and  c=  180°— C 

(Book  IX.  Prop.  VII.) :  hence,  omitting  the  ',  since  the  equa- 
tions   are    applicable  to  any  triangle,  we  shall  have 


cos 


l.^R^A"^  h  (A  +  B-C)  cos  i  (A  +  C-B)^ 
sin  B  sin  C 


cos  1  h=R^Ao^  i  (A  +  B-C)  cos  I  (B  +  C-A) 

sin  A  sin  C 

cos  I  c==R^/'cos  i  (A+C— B)  cos  j  (B  +  C— A) 

pin  A  sin  B. 


(6.) 


256 


SPHERICAL  TRIGONOMETRY. 


Putting  S=A  +  B  +  C,  we  shall  have 

JS— A=|(C  +  B--A),  JS— B=:J(A+C— B)  and 
^S— C=i(A+B--C),  hence 


cosi 


sin  B  sin  C 


cos  U=R^A'^^  gS— C)  cos  qS— A) 
sin  A  sin  C 


cos 


,^^^^^^/cos  gS-B)  cos  (xS 


-A) 


sin  A  sin  B 


HI-) 


VII.  If  we  apply  equations  (2.)  to  the  polar  triangle,  we 
shall  have 

— R2  cos  A'=R  cos  B'  cos  C — sin  B'  sin  C  cos  a\ 

Or,  omitting  the  ',  since  the  equation  is  applicable  to  any  tri- 
angle, we  have  the  three  symmetrical  equations, 

R^.cos  A=sin  B  sin  C  cos  a — R  cos  B  cos  C  \ 
R^.cos  B=sin  A  sin  C  cos  b — R  cos  A  cos  C  >  (8.) 
R^.cos  C=sin  A  sin  B  cos  c — R  cos  A  cos  B  / 

That  is,  radius  square  into  the  cosine  of  either  angle  of  a  sphe- 
rical  triangle,  is  equal  to  the  rectangle  of  the  sines  of  the  two  other 
angles  into  the  cosine  of  their  included  side,  minus  radius  into  the 
rectangle  of  their  cosines. 

VIII.  All  the  formulas  necessary  for  the  solution  of  spheri- 
cal triangles,  may  be  deduced  from  equations  marked  (2.).  If 
we  substitute  for  cos  b  in  the  third  equation,  its  value  taken 
from  the  second,  and  substitute  for  cos^  a  its  value  R^ — sin^  a, 
and  then  divide  by  the  common  factor  R.sin  a,  we  shall  have 

R.COS  c  sin  flnsin  c  cos  a  cos  B  + R.sin  b  cos  C. 

T,      '        ,.      /,  X    .         .    ,     sin  B  sin  c 
But  equation  (1.)  gives  sin  6= -. — — —  ; 

hence,  by  substitution, 

R  cos  c  sin  a=sin  c  cos  a  cos  B  +  R. 

Dividing  by  sin  c,  we  have 

n  cose    .  T>  ,  T> 

R  -; —  sm  a=cos  a  cos  B  +  R 
sin  c 


sin  B  cos  C  sin  c 
sin  C 

sin  B  cos  C 


sin  C 


( 


SPHERICAL  TRIGONOMETRY.  257 

But,  ^J-^  (Art.  XVII.). 
Sin      It 

Tlierefore,        cot  c  sin  a=cos  a  cos  B  +  cot  C  sin  U. 

Hence,  we  may  write  the  three  symmetrical  equations, 

cot  a  sin  &=cos  b  cos  C  +  cot  A  sin  C  n 
cot  b  sin  c=cos  c  cos  A  +  cot  B  sin  A  >  (9.) 
cot  c  sin  arzco's  a  cos  B  +  cot  C  sin  B  / 

That  is,  in  every  spherical  triangle,  the  cotangent  of  one  of  the 
sides  into  the  sine  of  a  second  side,  is  equal  to  the  cosine  of  the  se- 
cond side  into  the  cosine  of  the  included  angle,  plus  the  cotangent 
of  the  angle  opposite  the  first  side  into  the  sine  of  the  included 
angle. 

IX.  We  shall  terminate  these  formulas  by  demonstrating 
J^apier's  Analogies,  which  serve  to  simplify  several  cases  in  the 
;    solution  of  spherical  triangles. 

If  from  the  first  and  third  of  equations  (2.),  cos  c  be  elimi- 
nated, there  will  result,  after  a  little  reduction, 

R  cos  A  sin  c=R  cos  a  sin  b — cos  C  sin  a  cos  b. 

By  a  simple  pernlutation,  this  gives 

R  cos  B  sin  c=R  cos  b  sin  a — cos  C  sin  b  cos  a. 

Hence  by  adding  these  two  equations,  and  reducing,  we  shall 
have 

sin  c  (cos  A  +  cos  B)=(R — cos  C)  sin  (a +6) 

.  sin   c     sin  a     sin  6 

But  smce     - — 7^=-^^ — t  =  - — 5,  we  shall  have 
sm  C     sm  A     sm  B 

sin  c  (sin  A'+sin  B)=sin  C  (sin  a+sin  6),    and 

sin  c  (sin  A — sin  B)=sin  C  (sin  a — sin  b). 

Dividing  these  two  equations  successively  by  the  preceding 
one  ;  we  shall  have 

sin  A  +  sin  B_     sin  C         sin  cK  +  sin  b 
cos  A  +  cos  B~R — cos  C  '  sin  {a-\-b) 
sin  A — sin  B_     sin  C         sin  a — sin  b 
cos  A+cos  B~R — cos  C  *     sin  (a +6)* 

Y* 


258  V  SPHERICAL  TRIGONOMETRY, 

And  reducing  these  by  the  formulas  in  Articles  XXIII.  and 
XXIV.,  there  will  result  -^ 

ta„gJ(A+B)=cotiC.^-^ii)^  \ 

cos^(a+i)  jifj^ 

tangi(A-B)=cotJC.t"4^. 
-^     Sin  J  (a  +  6) 

Hence,  two  sides  a  and  h  with  the  included  angle  C  being 
given,  the  two  other  angles  A  and  B  may  be  found  by  the 
analogies, 

cos^-(a+i)  :  cosJ(a! — h)  :  :  cot^C  :  tangJ(A+B) 
sin  J  (rt  +  6)  :  sin  \  (a — b)  :  :  cot  |  C  :  tang  J  (A — B). 
If  these  same  analogies   are  applied  to  the  polar  triangle  of 
ABC,  we  shall  have  toput  180°— A',  180°— B',  180°— a',  180°- 6',  • 
180° — c',  instead  of  a,  6,  A,  B,  C,  respectively;  and  for  the  result, 
we  shall  have  after  omitting  the  ',  these  two  analogies, 

cosJ(A  +  B)  :  cosJ(A — B)  :  :  tangjc  :  tangJ(flr+6) 
sinJ(A  +  B)  :  sin^(A — B)  :  :  tangjc  :  tang  J  (a — J), 
by  means  of  which,  when  a  side  c  and  the  two  adjacent  angles.^ 
A  and  B  are  given,  W3  are  enabled  to  find  the  two  other  sides 
a  and  b.     These  four  proportions  are  known  by  the  name  of 
Napier^s  Analogies, 

X.  In  the  case  in  which  there  are  given  two  sides  and  an 
angle  opposite  one  of  them,  there  will  in  general  be  two  solu- 
tions corresponding  to  the  two  results  in  Case  II.  of  rectilineal 
triangles.  It  is  also  plain  that  this  ambiguity  will  extend  itselt 
to  the  corresponding  case  of  the  polar  triangle,  that  is,  to  the 
case  in  which  there  are  given  tvv^o  angles  and  a  side  opposite 
one  of  them.  In  every  case  we  shall  avoid  all  false  solutions 
by  recqllecting, 

1st.  That  every  angle,  and  every  side  of  a  spherical  triangle 
is  less  than  180°. 

2d.  That  the  greater  angle  lies  opposite  the  greater  side,  and 
the  least  angle  opposite  the  least  side,  and  reciprocally. 


NAPIER'S  CIRCULAR  PARTS. 

XI.  Besides  the  analogies  of  Napier  already  demonstrated, 
that  Geometer  also  invented  rules  for  the  solution  of  all  the 
cases  of  right  angled  spherical  triangles. 


fl^         ^  "Spherical  TRIGONOMETRY.  259 

In  every  right  angled  spherical 
triangle  BAG,  there  are  six  parts : 
three  sides  and  three  angles.  If 
we  omit  the  consideration  of  the 
right  angle,  which  is  always 
known,  there  will  be  five  remain- 
ing parts,  two    of  which   must 

.     be  given  before  the  others  can 

I     be  determined. 

f         The  circular  parts,  as  they  are  called,  arp  the  two  sides  c  and  h, 

i  about  the  right  angle,  the  complements  of  the  oblique  angles  B 
and  G,  and  the  complement  of  the  hypothenuse  a.  Hence  there 
are  five  circular  parts.  The  right  angle  A  not  being  a  circular 
part,  is  supposed  not  to  separate  the  circular  parts  c  and  6,  so 
that  these  parts  are  considered  as  adjacent  to  each  other. 

If  any  two  parts  of  the  triangle  be  given,  their  corresponding 
circular  parts  will  also  be  known,  and  these  together  with  a 
required  part,  will  make  three  parts  under  consideration.  Now, 
these  three  parts  will  all  lie  together,  or  one  of  them  will  he  sepa- 
rated from  both  of  the  others.  For  example,  if  B  and  c  were 
given,  and  a  required,  the  three  parts  considered  would  lie 
together.  But  if  B  and  G  were  given,  and  h  required,  the  parts 
would  not  lie  together ;  for,  B  would  be  separated  from  C  by 
the  part  a,  and  from  h  by  the  part  c.  In  either  case  B  is  the 
middle  part.  Hence,  when  there  are  three  of  the  circular  parts 
under  consideration,  the  middle  part  is  that  one  of  them  to  which 
both  of  the  others  are  adjacent,  or  from  which  both  of  them  are 
separated.  In  the  former  case  the  parts  are  said  to  be  adjacent, 

*'     and  in  the  latter  case  the  parts  are  said  to  be  opposite. 

This  being  premised,  we  are  now  to  prove  the  following 
rules  for  the  solution  of  right  angled  spherical  triangles,  which 
it  must  be  remembered  apply  to  the  circular  parts,  as  already 
defined. 

1st.  Radius  into  the  sine  of  the  middle  part  is  equal  to  the  rect- 
angle of  the  tangents  of  the  adjacent  paints. 

2d.  Radius  into  the  sine  of  the  middle  part  is  equal  to  the  rect- 
angle of  the  cosines  of  the  opposite  parts. 

These  rules  are  proved  by  assuming  each  of  the  five  circu- 
lar parts,  in  succession,  as  the  middle  part,  and  by  taking  the 
extremes  first  opposite,  then  adjacent.  Having  thus  fixed  the 
three  parts  which  are  to  be  considered,  take  that  one  of  the 
general  equatigns  for  oblique  angled  triangles,  which  shall  con  • 
tain  the  three  corresponding  parts  of  the  triangle,  together  with 
the  right  angle :  then  make  A  =  90^,  and  after  making  the  reduc- 
tions corresponding  to  this  supposition,  the  resulting  equation 
will  prove  the  rule  for  that  particular  case. 


260  SPHERICAL  TRIGONOMETRY. 

For  example,  let  comp.  a  be  the  middle  part  and  the  ex- 
tremes opposite.  The  equation  to  be  applied  in  this  case  must 
contain  a,  b,  c,  and  A.  The  first  of  equations  (2.)  contains  these 
four  quantities :  hence 

R^  cos  a=R  cos  b  cos  c+sin  b  sin  c  cos  A. 

If  A=90°  cos  AznO ;  hence 

Rcos  fl!=cos  6cosc; 

that  is,  radius  into  the  sine  of  the  middle  part,  (which  is  the 
complement  of  a,)  is  equal  to  the  rectangle  of  the  cosines  of  the 
opposite  parts. 

Suppose  now  that  the  complement 
of  a  were  the  middle  part  and  the  ex- 
tremes adjacent.  The  equation  to  be 
applied  must  contain  the  four  quan- 
tities fif,  B,  C,  and  A.     It  is  the  first 

of  equations  (8.). 

c 

R^  cos  A:=sin  B  sin  C  cos  a — R  cos  B  cos  C. 

Making  A =90°,  we  have 

sin  B  sin  C  cos  fl=R  cos  B  cos  C,    or 

R  cos  a=cot  B  cot  C ; 

that  is,  radius  into  the  sine  of  the  middle  part  is  equal  to  the 
rectangle  of  the  tangent  of  the  complement  -of  B  into  the  tan- 
gent of  the  complement  of  C,  that  is,  to  the  rectangle  of  the 
tangents  of  the  adjacent  circular  parts. 

Let  us  now  take  the  comp.  B,  for  the  middle  part  and  the 
extremes  opposite.  The  two  other  parts  under  consideration 
will  then  be  the  perpendicular  b  and  the  angle  C.  The  equation 
to  be  applied  must  contain  the  four  parts  A,  B,  C,  and  6  :  it  is  the 
second  of  equations  (8.), 

R"  cos  B=sin  A  sin  C  cos  b — R  cos  A  cos  C. 
Making  A = 90°,  we  have,  after  dividing  by  R, 

R  cos  B  =  sin  C  cos  b. 
Let  comp.  B  be  still  the  middle  part  and  the  extremes  adja- 
cent.    The  equation  to  be  applied  must  then  contain  the  four 
four  parts  a,  B,  c,  and  A.     It  is  similar  to  equations  (9.). 

cot  a  sin  c=cos  c  cos  B  4- cot  A  sin  B 
But  if  A =90°,  cot  A =0 ;  hence, 

cot  a  sin  cncos  c  cos  B  ;     or  . 
R  cos  B=:cot  a  tang  c. 


SPHERICAL  TRIGONOMETRY.  261 

And  by  pursuing  the  same  method  of  demonstration  when  each 
circular  part  is  made  the  middle  part,  we  obtain  the  five  fol- 
lowing equations,  which  embrace  all  the  cases, 

R  cos  a=cos  b  cos  c=cot  B  cot  C" 

R*  cos  Brrcos  b  sin  C=cot  a  tang  c 

R  cos  C=cos  csinB=cot  a  tang  b  Y  O-^') 

R  sin  Z>=sinasinB=tangccotC 

R  sin  c=sin<2sinC=tang6cotB> 

We  see  from  these  equations  that,  if  the  middle  part  is  required 
we  must  begin  the  proportion  with  radius  ;  and  when  one  of  the 
extremes  is  required  we  must  begin  the  proportion  with  the  other 
extreme. 

We  also  conclude,  from  the  first  of  the  equations,  that  when 
the  hy  pothenuse  is  less  than  90°,  the  sides  b  and  c  will  be  of  the  same 
species,  and  also  that  the  angles  B  and  C  will  likewise  be  of  the 
same  sjiecies.  When  a  is  greater  than  90°,  the  sides  b  and  c  will 
be  of  different  species,  and  the  same  will  be  true  of  the  angles  B 
and  C.  We  also  see  from  the  two  last  equations  that  a  side  and 
its  opposite  angle  will  always  be  of  the  same  species. 

These  properties  are  proved  by  considering  the  algebraic 
signs  which  have  been  attributed  to  the  trigonometrical  lines, 
and  by  remembering  that  the  two  members  of  an  equation  must 
always  have  the  same  algebraic  sign. 


SOfejJTION  OF  RIGHT    ANGLED   SPHERICAL  TRIANGLES    BY 
LOGARITHMS. 

It  is  to  be  observed,  that  when  any  element  is  discovered  in 
the  form  of  its  sine  only,  there  may  be  two  values  for  this  ele- 
ment, and  consequently  two  triangles  that  will  satisfy  the  ques- 
tion ;  because,  the  same  sine  which  corresponds  to  an  angle  or 
an  arc,  corresponds  likewise  to  its  supplement.  This  will  not 
take  place,  when  the  unknown  quantity  is  determined  by  means 
of  its  cosine,  its  tangent,  or  cotangent.  In  all  these  cases,  the 
sign  will  enable  us  to  decide  whether  the  element  in  question  is 
less  or  greater  than  90° ;  the  element  will  be  less  than  90°,  if  its 
cosine,  tangent,  or  cotangent,  has  the  sign  +  ;  it  will  be  greater 
if  one  of  these  quantities  has  the  sign  — . 

In  order  to  discover  the  species  of  the  required  element  of 
the  triangle,  we  shall  annex  the  minus  sign  to  the  logarithms  of 
all  the  elements  whose  cosines,  tangents,  or  cotangents,  are 
negative.     Then  by  recollecting  that  the  product  of  the  two 


262  SPHERICAL  TRIGONOMETRY.         v , 

extremes  has  the  same  sign  as  that  of  the  means,  we  can  at  once 
determine  the  sign  which  is  to  be  given  to  the  required  element* 
and  then  its  species  will  be  known. 


EXAMPLES. 

1.  In  the  right  angled  spherical  tri- 
angle BAG,  right  angled  at  A,  there 
are  given  «— 64°  40' and  6=42°  12': 
required  the  remaining  parts. 

First,  to  find  the  side  c. 

c 

The  hypothenuse  a  corresponds  to  the  middle  part,  and  the 
extremes  are  opposite :  hence 

R  cos  «=cos  h  cos  c,    or 

As  cos       h  42°  12'         ar.-comp.       .log.  0.130296 

Is  to          R  -         -         -    •     -  -         -  10.000000 

So  is  cos    a  64^'  40'         -         -  -         -  9.631326 

Tocos        c  54°  43' 07"            -  -      '  -.  9.761622 

To  find  the  angle  B. '  • 

The  side  h  will  be  the  middle  part  and  the  extremes  oppo- 
site :  hence 

R  sin  6=cos  (comp.  a)  x  cos  (comp.  B)=sin  a  sin  B. 

As  sin      a  64°  40'         ar.-comp.        log.  0.043911 

I§  to  sin  h  42°  12'         -         -        -         -  9.827189 

Sois       R        - 10.000000 

To  sin     B  48°  00'  14"  ...         -  9.871100 


To  find  the  angle  C^ 

The  angle  C  is  the  middle  part  and  the  extremes  adjacent ; 
hence 

R  cos  C=cot  a  tang  &. 

As              R  .-•               ar.-comp.  log.  0.000000 

Is  to  cot      a  64°  40'         -         -         -  -  '  9.675237 

So  is  tang  6  .42°  12'         ...  -  9.957485 

To  cos        C  64°  34'  46;'  -         -         -  -  9.632722 

2.  In  a  right  angled  triangle  BAG,  there  are  given  the  hy- 
pothenuse a=105°  34',  and  the  angle  B=80°  40' :  required  the 
remaining  parts. 


SPHERICAL  TRIGONOMETRY.  2G3 


To  find  the  angle  C. 

The  hypothenuse 
jacent :  hence, 

will  be  the  middle  part 

and  the  extremes 

R  cos  a=cot  B  cot  C. 

As  cot 
Is  to  cos 

So  is 

B 
a 
R 

80 
105 

^  40'        ar.-comp.     log. 
^34'    - 

0.784220  + 

9.428717— 

10.000000  + 

To  cot 

C 

148 

=  30' 54"      - 

10.212937— 

Since  the  cotangent  of  C  is  negative  the  angle  C  is  greater 
than  90°,  and  is  the  supplement  of  the  arc  which  would  corres- 
pond to  the  cotangent,  if  it  were  positive. 

To  find  the  side  c. 
The  angle  B  will  correspond  to  the  middle  part,  and  the 
extremes  will  be  adjacent :  hence, 

R  cos  B=cot  a  tang  c. 

Ascot       a  105°  34'         ar.-comp.         log.  0.555053 — 

Is  to  R  10.000000  + 

So  is  cos  B  80°  40'  .         -         -         .  9.209992  + 

To  tang      c  149°  47' 36"  -         -         -  9/765045-^ 

To  find  the  side  h. 

The  side  h  will  be  the  middle  part  and  the  extremes  oppo- 
site: hence, 

R  sin  6= sin  a  sin  B. 

As  R       -        ar.  comp.        log.        -  0.000000 

To  sin       a  105°  34'  -         .         -         .  9.983770 

So  is  sin  B     80°  40'  -         -         -         .  9.994212 

To  sin        h  71°54' 33"       ....  9.977982 


OF  QUADRANTAL  TRIANGLES. 

A  quadrantal  spherical  triangle  is  one  which  has  one  of  its 
sides  equal  to  90°. 

Let  BAG  be  a  quadrantal  triangle 
in  which  the  side  a =90°.  If  we  pass 
to  the  corresponding  polar  triangle, 
we  shall  have  A' =  180°— a =90°,  B'  = 
180°— &,  G'  =  180°— c,  a'  =  180°— A, 
5'=180°— B,c'  =  180°— C;  from  which 
we  see,  that  the  polar  triangle  will  be 


264  SPHERICAL  TRIGONOMETRY.     . 

-#    , 
right  angled  at  A',  and  hence  every  case  may  be  referred  to 
a  right  angled  triangle. 

But  we  can  solve  the  quadrantal  triangle  by  means  of  the 
right  angled  triangle  in  a  manner  still  more  simple. 

In  the  quadrantal  triangle  BAG,  C 

in  which  BC  =  90°,  produce  the  side  A 

CA  till  CD  is  equal  to  90°,  and  con-  /   \ 

ceive  the  arc  of  a  great  circle  to  be  y  \  , 

drawn  through  B  and  D.    Then  C  ^^  \ 

will  be  the  pole  of  the  arc  BD,  and         ^^^^^  _^  A 

the  angle  C  will  be  measured  by   B^-s^Cl^ \ 

BD  (Book  IX.  Prop.  VI.),  and  the  \..,       ^  \b 

angles  CBD  and  D  will  be  right  an-  "i^'---..      / 

gles.     Now  before  the  remaining  "D 

parts  of  the  quadrantal  triangle  can 

be  found,  at  least  two  parts  must  be  given  in  addition  to  the 
side  BC  =  90° ;  in  which  case  two  parts  of  the  right  angled  tri- 
angle BDA,  together  with  the  right  angle,  become  known. 
Hence  the  conditions  which  enable  us  to  determine  one  of  these 
triangles,  will  enable  us  also  to  determine  the  other. 

3.  In  the  quadrantal  triangle  BCA,  there  are  given  CB=r50°, 
the  angle  C=42°  12',  and  the  angle  A=115°  20' :  required  the 
remaining  parts. 

Having  produced  CA  to  D,  making  CD =90°  and  drawn  the 
arc  BD,  there  will  then  be  given  in  the  right  angled  triangle 
BAD,  the  side  «=C=42°  12',  and  the  angle  BAD=180°— 
BAC  =  180°— 115°  20'=64°40',to  find  the  remaining  parts. 

To  find  the  side  d. 

The  side  a  will  be  the  middle  part,  and  the  extremes  oppo- 
site: hence, 

R  sin  a=sinA  sin  d. 


As  sin     A 
Is  to        R 
So  is  sin  a 

64°  40' 
42°  12' 

ar.-comp. 

log.     0.043911 

10.000000 

9.827189 

To  sin      d 

48°  00'  14" 

■ 

9.871100 

To  find  the  angle  B. 

The  angle  A  will  correspond  to  the  middle  part,  and  the  ex- 
tremes  will  be  opposite :  hence 

R  cos  A=sin  B  cos  a. 


As  cos      a 
Is  to        R 
So  is  cos  A 

42°  12'          ar.-comp. 
64°  40' 

log. 

0.130^96 

10.000000 

9.631326 

To  sin      B 

35°  16'  53" 

. 

9.761622 

SPHERICAL  TRIGONOMETRY.  265 

To  find  the  side  b: 

The  side  b  will  be  the  middle  part,  and  the  extremes  adja- 
cent :  hence, 


R  sin  6= cot  A  tang  a. 

As            R 

ar.-comp.        log. 

0.000000 

Is  to  cot  A 

64°  40'          .        -        -        . 

9.675237 

So  is  tang  a 

42°  12'          -         .        ..        . 

9.957485 

To  sin       6 

25°  25' 14" 

9.632722 

Hence,     CA=90°— 6=90°— 25°  25'  14"  =64°  34'  46" 

CBA=:90°— ABD  =  90°— 35°  16' 53" =54°  43  07" 
BA=^  .         .         -         .         =z48°00'15". 

4.  In  the  right  angled  triangle  BAC,  right  angled  at  A,  there 
are  given  <z=115°  25',  and  c=60°  59' :  required  the  remaining 
parts. 

(  Bzr:148°56'45" 

Ans.    )  C=  75°  30'  33" 

(  6  =152°  13' 50". 

5.:  In  the  right  angled  spherical  triangle  BAC,  right  angled 
at  A,  there  are  given  c= 116°  30' 43",  and  6=29°  41'  32":  re- 
quired the  remaining  parts. 

(  C=:103°  52' 46" 

^715.   )  B=  32°  30'  22" 

(a  =112°  48' 58". 

6.  In  a  quadrantal  triangle,  there  are  given  the  quadrantal 

side  —90°,  an  adjacent  side  =115°  09',  and  the  included  angle 

=  115°  55' :  required  the  remaining  parts. 

(  side,  113°  18' 19" 

^^^'    \  or.„lo.    i  117°  33' 52" 
)  angles,  ^  j^^,  ^^,  ^^,,^ 


SOLUTION  OF  OBLIQUE  ANGLED  TRIANGLES  BY  LOGARITHMS 

There  are  six  cases  which  occur  in  the  solution  of  oblique 
angled  spherical  triangles. 

1.  Having  given  two  sides,  and  an  angle  opposite  one  of 
them. 

2.  Having  given  two  angles,  and   a  side  opposite  one  o( 
them. 

3.  Having  given  the  three  sides  of  a  triangle,  to  &<t  tS^ 
•angles. 


266  SPHERICAL  TRIGONOMETRY. 

4.  Having  given  the  three  angles  of  a  triangle,  to  find  the 
sides. 

5.  Having  given  two  sides  and  the  included  angle. 

6.  Having  given  two  angles  and  the  included  side. 


CASE  I. 

Given  two  sides,  and  an  angle  opposite  one  of  them,  to  find  the  re- 
maining parts. 

For  this  case  we  employ  equation  (1.)  ; 

As  sin  a  :  sin  5  :  :  sin  A  :  sin  B. 

Ex.  1.  Given  the  side  fl=44° 
13'  45",  6=:84°  14'  29"  and  the 
angle  A=32°  26'  07"  :  required 
the  remaining  parts. 

To  find  the  angle  B. 

As  sin       a  44°  13' 45"      ar.-comp.         log.  0.156437 

Is  to  sin    h  84°  14' 29"  -         -         -  9.997803 

So  is  sin  A  32°  26'  07"  -         -         -  9.720445 

To  sin      B  49°  54' 38"  or  sin  E' 130°  5' 22"  9.883685 

Since  the  sine  of  an  arc  is  the  same  as  the  sine  of  its  supple- 
ment, there  will  be  two  angles  corresponding  to  the  logarithmic 
sine  9.883685  and  these  angles  will'  be  supplements  of  each 
other.  It  does  not  follow  however  that  both  of  them  will  satisfy 
all  the  other  conditions  of  the  question.  If  they  do,  there  will 
be  two  triangles  ACB',  ACB  ;  if  not,  there  will  be  but  one. 

To  determine  the  circumstances  under  which  this  ambiguity 
arises,  we  will  consider  the  2d  of  equations  (2.). 

R^  cos  6=R  cos  a  cos  c+sin  a  sin  c  cos  B. 

from  which  we  obtain 

_     R^  cos  h — R  cos  a  cos  c 

C0SB  = :— : ; .  ' 

sm  a  sm  c 

Now  if  cos  h  be  greater  than  cos  a,  we  shall  have  . 

R^  cos  fe>R  cos  a  cos  c, 

or  the  sign  of  the  second  member  of  the  equation  will  depend 
on  that  of  cos  6.    Hence  cos  B  and  cos  h  will  have  the  same 


SPHERICAL  TRIGONOMETRY.  2G7 

sign,  or  B  and  h  will  be  of  the  same  species,  and  there  will  be 
but  one  triangle. 

But  when     cos  &>cos  a,  sin  6<sin  a  :  hence, 

If  the  sine  of  the  side  opposite  the  required  angle  he  less  than 
the  sine  of  the  other  given  side,  there  will  he  hut  one  triangle. 

If  however,  sin  6>sin  a,  the  cos  h  will  be  less  than  cos  a, 
and  it  is  plain  that  such  a  value  may  then  be  given  to  c  as  to 
render 

R^  cos  &  <  R  cos  a  cos  c, 

or  the  sign  of  the  second  member  may  be  made  to  depend  on 
cos  c. 

We  can  therefore  give  such  values  to  c  as  to  satisfy  the  two 
equations 

_,     R^  cos  h — R  cos  a  cos  c 
H-cos  B= — 


;os  B  = 


sm  a  sm  c 

R^  cos  h — R  cos  a  cos  c 

sin  a  sin  c 


Hence,  if  the  sine  of  the  side  opposite  the  required  angle  he 
greater  than  the  sine  of  the  other  given  side,  there  will  he  two  tri- 
angles which  will  fulfil  the  given  conditions. 

Let  us,  however,  consider  the  triangle  ACB,  in  which  we  are 
yet  to  find  the  base  AB  and  the  angle  C.  We  can  find  these 
parts  most  readily  by  dividing  the  triangle  into  two  right  angled 
triangles.  Draw  the  arc  CD  perpendicular  to  the  base  AB  : 
then  in  each  of  the  triangles  there  will  be  given  the  hypothe- 
nuse  and  the  angle  at  the  base.  And  generally,  when  it  is 
proposed  to  solve  an  oblique  angled  triangle  by  means  of  the 
right  angled  triangle,  we  must  so  draw  the  perpendicular  that 
it  shall  pass  through  the  extremity  of  a  given  side,  and  lie  oppo- 
site to  a  given  angle. 

To  find  the  angle  C,  in  the  triangle  ACD. 

As  cot          A  32°  26'  07"     ar.-comp.       log.  9.803105 

Is  to             R 10.000000 

So  is  cos       6  84'^  14'  29"          -         -         -  9.001465 

To  cot    ACD  86^  21' 06"          -         -         -  ¥.804570 

To  find  tlio  angle  C  in  the  triangle  DCB. 

As  cot           B  49°  54'  38"     ar.-comp.     log.  0.074810 

Is  to              R 10.000000 

So  is  cos       a  44°  13' 45"      -         -         -  9.855250 

To  cot     DCB  49°  35' 38"      -         -         -  9.930060 

t      Hence  ACB=135o  56'  47'^ 


^08  SPHERICAL  TRIGONOMETRY. 

To  find  the  side  AB. 

As  sin  A     32°  26'  07"  ar.-comp.  log.  0.270555 

Is  to  sin  C  135°  56' 47"  -         -         -  9.842191 

So  is  sin  a     44°  13' 45"  -         -         -  9.843563 

Tosin  c  115°  16' 29"  -         -         -  9.956309 

The  arc  64°  43'  31",  which  corresponds  to  sin  c  is  not  the 
value  of  the  side  AB  :  for  the  side  AB  must  be  greater  than  6, 
since  it  lies  opposite  to  a  greater  angle.  But  6  =  84°  14'  29"  : 
hence  the  side  AB  must  be  the  supplement  of  64°  43'  31",  or 
115°  16' 29". 

Ex.  2.  Given  &=91°  03'  25",  a=40°  36'  37",  and  A=35°  57' 
15":  required  the  remaining  parts,  when  the  obtuse  angle  B  is 
taken. 

(  B  =  115°35'41" 

^715.    )  C=   58°  30' 57" 
/  c  =  70°  58'  52" 


CASE  II. 

Having  given  two  angles  and  a  side  opposite  one  oftliem^  to  find 
the  remaining  parts. 

For  this  case,  we  employ  the  equation  (l.^ 
sin  A  :  sin  B  :  :  sin  a  :  sin  h. 

Ex.  1.  In  a  spherical  triangle  ABC,  there  are  given  the  angle 
A=:50°  12',  B  =  58°  8',  and  the  side  a=62°  42' ;  to  find  the  re- 
maining parts. 

To  find  the  side  h. 

As  sin       A  50°  12'         ar.-comp.         log.  0.114478 

Is  to  sin    B  58°  08'         -         -         -         -  9.929050 

So  is  sin    a  62°  42'  -         -         -         -  9.948715 

To  sin        h  79°  12'  10",  or  100°  47'  50"  9.992243 

We  see  here,  as  in  the  last  example,  that  there  are  two  arcs 
corresponding  to  the  4th  term  of  the  proportion,  and  these  arcs 
are  supplements  of  each  other,  since  they  have  the  same  sine. 
It  does  not  follow,  however,  that  both  of  ih^m  will  satisfy  all 
the  conditions  of  the  question.  If  they  do,  there  will  be  two 
liiangles ;  if  not,  there  will  be  but  one. 


SPHERICAI.  TRIGONOMETRY.  269 

To  determine  when  there  are  two  triangles,  and  also  when 
there  is  but  one,  let  us  consider  the  second  of  equations  (8.) 

R^  cos  B=sin  A  sin  C  cos  b — R  cos  A  cos  C,      which  gives 

.     R^cosB  +  R  cos  AcosC 

cos  0=  = A = Ti . 

sm  A  sin  C 

Now,  if  cos  B  be  greater  than  cos  A  we  shall  have 

W  cos  B  >R  cos  A  cos  C, 

and- hence  the  sign  of  the  second  member  of  the  equation  will 
depend  on  that  of  cos  B,  and  consequently  cos  b  and  cos  B  will 
have  the  same  algebraic  sign,  or  b  and  B  will  be  of  the  same 
species.     But  when  cos  B  >cos  A  the  sin  B<sin  A  :  hence 

If  the  sine  of  the  angle  opposite  the  required  side  be  less  than 
the  sine  of  the  other  given  angle,  there  will  be  but  one  solution. 

If,  however,  sin  B>sin  A,  the  cos  B  will  be  less  than  cos  A, 
and  it  is  plain  that  such  a  value  may  then  be  given  to  cos  C,  as 
to  render 

R^cos  B<R  cos  A  cos  C, 

or  the  sign  of  the  second  member  of  the  equation  may  be  made 
to  depend  on  cos  C.  We  can  therefore  give  such  values  to  C 
as  to  satisfy  the  two  equations 

R^  cos  B  +  R  cos  A  cos  C 

+COS  6= -. — 7 — -. — Pi ,     and 

sm  A  sm  C 

R^  cos  B  +  R  cos  A  cos  C 

cos  0= : -r—. -^ . 

sm  A  sm  C 

Hence,  if  the  sine  of  the  angle  opposite  the  required  side  he 
greater  than  the  sine  of  the  other  given  angle  there  will  be  two 
solutions. 

Let  us  first  suppose  the  side  b  to  be  less  than  90°,  or  equa. 
to  79°  12'  10". 

If  now,  we  let  fall  from  the  angle  C  a  perpendicular  on  the 
base  BA,  the  triangle  will  be  divided  into  two  right  angled  tri- 
angles, in  each  of  which  there  will  be  two  parts  known  besides 
the  right  angle. 

Calculating  the  parts  by  Napier's  rules  we  find, 

C=130°54'28" 

c=119°03'26". 

If  we  take  the  side  6=100°  47'  60",  we  shall  find 

C=156°15'06" 

c^l52°  14'  18". 


270  SPHERICAL  TRIGONOMETRY. 

Ex.  2.  In  a  spherical  triangle  ABC  there  are  given  A=103° 
59'  57",  B=46°  18'  7",  and  a=42^  8'  48"  ;  required  the  remain- 
ing parts. 

There  will  but  one  triangle,  since  sin  B<sin  A. 

/  b  =30° 
Ans.    )  C=36°  7'  54'- 
}  c  =24°  3'  56". 


CASE  III. 

Having  given  the  three  sides  of  a  spherical  triangle  to  find  the 

angles. 


For  this  case  we  use  equations  (3.). 


/sin  J  5  sin  {^s — a) 
cos^A=RV  sin  6  sine 

Ex,  1.  In  an  oblique  angled  spherical  triangle  there  are 
given  fl=:56°  40',  6=83°  13'  and  c=114°  30' ;  required  the 
angles. 

l{a  +  h^c)=ls        =127°  11'  30" 
^(6  +  c— a)=(^5— a)=70°  31'  30". 

Log  sin      ^s  127°  11' 30"          -         -         -  9.901250 

log  sin  (|5— a)  70°  31' 30"           -         -         -  9.974413 

—log  sin     h     83°  13'               ar.-comp.  0.003051 

—log  sin     c  114°  30'                ar.-comp.  0.040977 

Sum 19.919691 

Half  sum  =log  cos  \k  24°  15',  39"     -         -  9.959845 

Hence,  angle  A=48°  31'  18". 

The  addition  of  twice  the  logarithm  of  radius,  or  20,  to  the 
numerator  of  the  quantity  under  the  radical  just  cancels  the  20 
which  is  to  be  subtracted  on  account  of  the  arithmetical  com- 
plements, to  that  the  20,  in  both  cases,  may  be  omitted. 

Applying  the  same  formulas  to  the  angles  B  and  C,  we  find, 

B=  62°  55'  46" 
C  =  125°  19' 02". 
Ex.  2.  In  a  spherical  triangle  there  are  given  arr40°  18'  29", 
6=67°  14'  28",  and  c~89°  47'  6"  :  required  the  three  angles. 

(  A=  34°  22'  16" 

Ans.    ;  Br=  53°  35'  16" 

'  {  C  =  119o  13' 32" 


SPHERICAL  TRIGONOMETRY. 


:27J 


CASE  IV. 

Having  given  the  three  angles  of  a  spherical  triangle,  to  find  the 
three  sides. 


For  this  case  we  employ  equations  (7.) 


^       ^./cos(^S-B)cos(^S-C) 

cos§a=Rv    T>        r ' 

sm  15  sin  O 


Ex.  1.  In  a  spherical  triangle  ABC  there  are  given  A=48° 
30',  B-=125°  20',  and  C  =  62°  54' ;  required  the  sides. 

J(A  +  B  +  C)  =  |S=   118°  22' 


(iS-A)        . 

=     69°  52' 

(iS-B)        - 

=_  6°  58' 

(^S-C)        - 

=    -650  28' 

Log  cos  (iS— B)  —6°  58' 

... 

9.996782 

log  cos    (JS— C)    55°  28' 

- 

9.753495 

—log  sin  "  B         125°  20' 

ar.-comp. 

0.088415 

—log  sin      C          62°  54' 

ar.-comp. 

0.050506 

Sura         -         -         -         - 

- 

19.889198 

Half  sum=log  cos  iA=28°  19' 

48" 

9.944599 

Hence,  side  a=56°  39'  36". 

In  a  similar  manner  we  find. 


6  =  1140  29'  58" 
c—  83°  12'  06". 


Ex.  2.  In  a  spherical  triangle  ABC,  there  are  given  A= 109° 
65'  42",  B  =  116°  38'  33",  and  0=120°  43'  37" ;  required  the 
three  sides. 


Ans. 


a=  98°  21.'  40" 
&=109°  50' 22" 
c  =  115°  13' 26" 


CASE  V. 


Having  given  in  a  spherical  triangle,  two  sides  and  their  in- 
cluded angle,  to  find  the  remaining  parts. 


272  SPHERICAL  TRIGONOMETRY. 

For  this  case  we  employ  the  two  first  of  Napier's  Analogies, 
cos  ^{a-{-b)  :  cos  i(a — b)  :  :  cot  iC  :  tang  i(A  +  B) 
sin  ^(a-\-b)  :  sin  ^{a—b)  :  :  cot  ^C  :  tang  i(A— B). 

Having  found  the  half  sum  and  the  half  diflference  of  the 
angles  A  and  B,  the  angles  themselves  become  known  ;  for,  the 
greater  angle  is  equal  to  the  half  sum  plus  the  half  difference, 
and  the  lesser  is  equal  to  the  half  sum  minus  the  half  diffe. 
rence. 

The  greater  angle  is  then  to  be  placed  opposite  the  greater 
side.  The  remaining  side  of  the  triangle  can  then  be  found  by 
Case  II. 

Ex.  1.  In  a  spherical  triangle  ABC,  there  are  given  a=68° 
46'  2",  b=ST  10',  and  0=39=^  23' ;  to  find  the  remaining  parts. 

^(a  +  &)  =  52°  58'  1",  i(a—b)  =  l5o  48'  1",  JC  =  19°41'  30". 

As  cos      !(«  +  &)  52°  58'    1"  log.  ar.-comp.  0.220210 

'     Is  to  cos    L(a—h)  15°  48'    1"  -  -         -  9.983271 

So  is  cot         iC    19°  41' 30"  -  -         -  10.446254 

Totangi{A  +  B)    77°  22' 25"  -  -         -  10.649735 

As  sin      ^(a  +  b)  52°  58'    1"  log.  ar.-comp.  0.097840 

Is  to  sin  i{a—b)  15°  48'    1"  -  -         -  9.435016 

So  is  cot  "iC      19°  41' 30"  -  -         -  10.446254 

Totangi(A— B)  43°37'21"  -  -         -  9.979110 

Hence,  A=77°  22'  25"  +  43°  37'  21"=120°  59'  46" 
B=77°  22'  25"— 430  37'  21"=  33°  45'  04" 
side  c         '         -         -         -  =  43°  37'  37". 

Ex,  2.  In  a  spherical  triangle  ABC,  there  are  given  6=:83'' 
19'  42':,  c=23°  27'  46",  the  contained  angle  A=20°  39'  48"; 
to  find  the  remaining  parts. 

(  B  =  156°  30'  16" 

Ans,    )C=     9°  11' 48" 

)  a=  61°  32'  12". 


CASE  VI. 

In  a  spherical  triangle^  having  given  two  angles  and  the  included 
side  to  find  the  remaining  parts. 


SPHERICAL  TRIGONOMETRY.  273 

For  this  case  we  employ  the  second  of  Napier's  Analogies, 
cos  J(A  +  B)  :  cos  J  (A — B)  :  :  tang  |c  :  tangj(a  +  fc) 
sin^-(A  +  B)  :  sin  J  (A — B)  :  :  tang  Jc  :  tang  J  (a — b). 

From  which  a  and  b  are  found  as  in  the  last  case.     The  re- 
maining angle  can  then  be  found  by  Case  I. 

Ex.  1.  In  a  spherical  triangle  ABC,  there  are  given  A= 81° 
38'  20",  B=70°  9'  38",  c=59°  16'  28"  ;  to  find  the  remaining 
parts. 

J(A  +  B)=75°  53'  59",l(A— B)=5°44'21",^c=29°  38'  11". 

As  cos       i{A  +  B)  75°  53' 59"   log.    ar.-comp.  0.613287 
Tocos        KA— B)     5°  44' 21"  -         -         9.997818 

So  is  tang       ic        29°  38' 11"  -        -         9.755051 


To  tang       i(a  +  6)  66°42'52"  -         -       10.366156 

As  sin        ^(A+B)  75°  53' 59"  log.  ar.-comp.  0.013286 
To  sin        4(A— B)     5°  14'  21"  -        -         9.000000 

So  is  tang       ^c         29°  38' 11"  -         -         9.755051 

To  tang       i{a—b)     3°  21'  25"  -         -         8.768337 

Hence       «=66o  42'  52"  +  3°  21'  25"=70°  04'  17'" 

6=66°  42'  52"-^3°  21'  25"=63°  21'  27" 

angle  C         -         -         -     =64°  46'  33". 

Ex,  2.  In  a  spherical  triangle  ABC,  there  are  given  A=34'* 
15'  3",  B=42°  15'  13",  and  0=76°  35'  36" ;  to  find  the  remain- 
ing  parts. 

(  a  =40°    0'  10" 
Ans,    H  =50°  10'  30" 
(C  =58°  23' 41". 


i   274  ) 


MENSURATION  OF  SURFACES. 

The  area,  or  content  of  a  surface,  is  determined  by  finding 
how  many  times  it  contains  some  other  surface  which  is  as- 
sumed as  the  unit  of  measure.  Thus,  when  we  say  that  a 
square  yard  contains  9  square  feet,  we  should  understand  that 
one  square  foot  is  taken  for  the  unit  of  measure,  and  that  this 
unit  is  contained  9  times  in  the  square  yard. 

The  most  convenient  unit  of  measure  for  a  surface,  is  a 
square  whose  side  is  the  hnear  unit  in  which  the  linear  dimen- 
sions of  the  figure  are  estimated.  Thus,  if  the  linear  dimen- 
sions are  feet,  it  will  be  most  convenient  to  express  the  area  in 
square  feet ;  if  the  linear  dimensions  are  yards,  it  will  be  most 
convenient  to  express  the  area  in  square  yards,  &c. 

We  have  already  seen  (Book  IV.  Prop.  IV.  Sch.),  that  the 
term,  rectangle  or  product  of  two  lines,  designates  the  rectan- 
gle constructed  on  the  lines  as  sides ;  and  that  the  numerical 
value  of  this  product  expresses  the  number  of  times  which  the 
rectangle  contains  its  unit  of  measure. 

PROBLEM  I. 

To  find  the  area  of  a  square,  a  rectangle,  or  a  parallelogram. 

Rule. — Multiply  the  base  by  the  altitude,  and  the  product  will 
be  the  area  (Book  IV.  Prop.  V.). 

1.  To  find  the  area  of  a  parallelogram,  the  base  being  12.25 
and  the  altitude  8.5.  Ans.  104.125. 

2.  What  is  the  area  of  a  square  whose  side  is  204.3  feet  1 

Ans.  41738.49  sq.ft. 

3.  What  is  the  content,  in  square  yards,  of  a  rectangle  whose 
base  is  66.3  feet,  and  altitude  33.3  feet?  Ans.  245.31. 

4.  To  find  the  area  of  a  rectangular  board,  whose  length  is 
12J-  feet,  and  breadth  9  inches.  Ans.  9f  sq.ft. 

5.  To  find  the  number  of  square  yards  of  painting  m  a  par- 
allelogram, whose  base  is  37  feet,  and  altitude  5  feet  3  inches 

Ms.  21yV 
PROBLEM  ir. 

To  find  the  area  of  a  triangle.     '* 

CASE  L 

When  the  base  and  altitude  are  given. 

Rule. — ^Multiply  the  base  by  the  aliiiude,  and  take  half  the 
product.  Or,  multiply  one  of  these  dimensions  by  half  the 
of/icr  (Book  IV.  Prop.  VI.). 


MENSURATION  OF  SURFACES.  275 

1.  To  find  the  area  of  a  triangle,  whose  base  is  625  and  alti- 
tude 520  feet.  Ans.   162500  sq.ft. 

2.  To  find  the  number  of  square  yards  in  a  triangle,  whose 
base  is  40  and  altitude  30  feet.  Ans,  66|. 

3.  To  find  the  number  of  square  yards  in  a  triangle,  whose 
base  is  49  and  altitude  25^  feet.  ^715.  68.7361. 

CASE  II. 
When  two  sides  and  their  included  angle  are  given. 

Rule. — Add  together  the  logarithms  of  the  two  sides  and  the 
logarithmic  sine  of  their  included  angle ;  from  this  sum  sub- 
tract the  logarithm  of  the  radius,  which  is  10,  and  the  remain- 
der will  he  the  logarithm  of  double  the  area  of  the  triangle. 
Find,  from  the  table,  the  number  answering  to  this  logarithm, 
and  divide  it  by  2 ;  the  quotient  will  be  the  required  area. 

Let  BAG  be  a  triangle,  in  which  there 
are  given  BA,  BC,  and  the  included  an- 
gleB. 

From  the  vertex  A  draw  AD,  perpen- 
dicular to  the  base  BC,  and  represent  the 
area  of  the  triangle  by  Q.     Then, 

R  :  sin  B  : :  BA  :  AD  (Trig 
hence,  ^p^BAxsinB 

R 

But,  Q^J^GxAD  (Book  IV.  Prop.  VI.) ; 
.^ 

hence,  by  substituting  for  AD  its  value,  we  have 

Q_BCxBAxsinB   ^^  QO_BCxBAxsin  B 
2R  R 

Taking  the  logarithms  of  both  numbers,  we  have 

log.  2Q=log.  BC  +  log.  BA  +  log.  sin  B— log.  R; 
which  proves  the  rule  as  enunciated. 

1.  What  is  the  area  of  a  triangle  whose  sides  are,  BC=: 
125.81,  BA=57.65,  and  the  included  angle  B  =  57°  25'? 

'  +log.  BC  125.81  ....  2.099715 
T-Uor.  i^„  on  J  +log.  BA  57.65  ....  1.760799 
Then,  log.  2Q=  ^    ^j^|  ^.^  g     ^^^  25' 9.925626 

—log.  R -     —10. 


log.  2Q     3.786140 

and  2Q=6111.4,    or  Q=: 3055.7,  the  required  area. 


276  MENSURATION  OF  SURFACES. 

2.  What  is  the  area  of  a  triangle  whose  sides  are  30  and  40, 
and  their  included  angle  28°  57' '(  Ans.  290.427. 

3.  What  is  the  number  of  square  yards  in  a  triangle  of  which 
the  sides  are  25  feet  and  21.25  feet,  and  their  included  angle 
45°?  .        .  Ans.  20.8694. 

CASE  III. 
When  the  three  sides  are  known. 

Rule. — 1.  Add  the  three  sides  together ,  and  take  half  their  sum. 

2.  From  this  half-sum  subtract  each  side  separately. 

3.  Multiply  together  the  half-sum  and  each  of  the  three  re- 
mainders, and  the  product  will  he  the  square  of  the  area  oj 
the  triangle.  Then,  extract  the  square  root  of  this  product, 
for  the  required  area. 

Or,  After  having  obtained  the  three  remainders,  add  together  the 
logarithm  of  the  half -sum  and  the  logarithms  of  the  respective 
remainders,  and  divide  their  sum  by  2 :  the  quotient  will  be 
the  logarithm  of  the  area. 

Let  ABC  be  the  given  triangle.  ^C 

Take  CD  equal  to  the  side  CB,  and 

draw  DB;  draw  AE  parallel  to  DB,  ly 

meeting  CB  produced,  in  E :  then         ,'''  /x  \y 

CE  will  be  equal  to  CA.     Draw       /  I>/-".^:  -v\B 

CFG  perpendicular  to  AE  and  DB,      /  i{y 

and  it  will  bisect  them  at  the  points      '. 

G  and  F.     Draw  FHI  parallel  to      k.<''  /^       J/ 

AB,  meeting  CA  in  H,  and  EA  pro-         '^  ,;A        yG    '   "^ 

duced,  in  I.     Lastly,  with  the  cen-  K' 

tre  H  and  radius  HF,  describe  the  circumference  of  a  circle, 
meeting  CA  produced  in  K:  this  circumference  will  pass 
through  I,  because  AI=FB=FD,  therefore,  HF=H1  ;  and  it 
will  also  pass  through  the  point  G,  because  FGI  is  a  right 
angle. 

Now,  since  HA=HD,  CH  is  equal  to  half  the  sum  of  the 
sides  CA,  CB ;  that  is,  CH=iCA+iCB;  and  since  HK  is 
equal  to  iIF=iAB,  it  follows  that 

CK=iAC  +  iCB  +  iAB=iS, 
by  representing  the  sum  of  the  sides  by  S. 

Again,  HK=HI=iIF=iAB,  or  KL=AB. 
Hence,  CL=CK-KL=iS— AB, 
and        AK=CK-  CA=|S— CA, 
and        AL=DK=CK— CD=iS— CB. 

Now,    AG  X  CG=  the  area  of  the  triangle  ACE, 
and  AG  x  FG=  the  area  of  the  triangle  ABE ; 

therefore,  AG  x  CF=:  the  area  of  the  triangle  ACB. 


MENSURATION  OF  SURFACES.  277 

Also,  by  similar  triangles, 

AG  :  CG  : :  I)F  :  CF,  or  AT  :  CF ; 

therefore,  AG  x  CF=:  triangle  ACB-CG  x  DF=CG  x  AI ; 
consequently,  AGxCFxCGx  AI=  square  of  the  area  ACB. 

But       CGxCF=CKxCL-JS(iS— AB), 
and  AGx  AI  =AKx  AL=(iS— CA)  x  (^S— CB) ; 

therefoie,AGxCFxCGxAI  -iS^S  — AB)  x  GS  — CA)  x 
QS — CB),  which  is  equal  to  the  square  of  the  area  of  the 
triangle  ACB. 

1.  To  find  the  area  of  a  triangle  whose  three  sides  are  20, 
30,  and  40. 

20  45  45  45  half-sum. 

W  20  30  40 

40  --  —  — 

.    —  25  1st  rem;  15  2d  rem.        5  3d  rem. 

2)90 

45  half-sum. 

Then,  45  X  25  X  15  X  5=84375. 
V  The  square  root  of  which  is  290.4737,  the  required  area. 

2.  How  many  square  yards  of  plastering  are  there  in  a  tri- 
angle whose  sides  are  30,  40,  and  50  feet  ?  Ans.  66|. 

PROBLEM  III. 

To  find  the  area  of  a  trapezoid. 

Rule. — Add  together  the  two  parallel  sides :  then  multiply  their 
sum  by  the  altitude  of  the  trapezoid,  and  half  the  product  will 
be  the  required  area  (Book  lY.  Prop.  VII.). 

1.  In  a  trapezoid  the  parallel  sides  are  750  and  1225,  and 
the  perpendicular  distance  between  them  is  1540 ;  what  is  the 
area?  Ans.  152075. 

2.  How  many  square  feet  are  contained  in  a  plank,  whose 
length  is  12  feet  6  inches,  the  breadth  at  the  greater  end  15 
inches,  and  at  the  less  end  11  inches?  Ans.  13|^|  sq.ft. 

3.  How  many  square  yards  are  there  in  a  trapezoid,  whose 
parallel  sides  are  240  feet,  320  feet,  and  altitude  66  feet  ? 

Ans.  2053^. 

PROBLEM  IV. 
To  find  the  area  of  a  quadrilateral. 

Rule. — Join  two  of  the  angles  by  a  diagonal,  dividing  the  quad* 
rilateral  into  two  triangles.  Then,  from  each  of  the  other 
angles  let  fall  a  'perpendicular  on  the  diagonal :  then  multiply 

A  a 


218 


MENSURATIC»N  OF  SURFACES. 


the  diagonal  by  half  the  sum  of  the  two  perpendiculars^  and 
the  product  wid  be  the  area. 


1.  What  is  the  area  of  the  quad- 
rilateral  ABCD,  the   diagonal    AC 
being   42,    and   the    perpendiculars 
D^,  B&,  equal  to  18  and  16  feet  ? 
Ans.  714. 


2.  How  many  square  yards  of  paving  are  there  in  the  quad- 
rilateral whose  diagonal  is  65  feet,  and  the  two  perpendiculars 


let  fall  on  it  28  and  33^  feet  ? 


Ans.  222J 


PROBLEM  V. 
To  find  the  area  of  an  irregular  polygon. 

Rule. — Draw  diagonals  dividing  the  proposed  polygon  into 
trapezoids  and  triangles.  Then  find  the  areas  of  these 
figures  separately  J  and  add  them  together  for  the  content  of 
the  whole  polygon. 

1.  Let  it  be  required  to  determine 
the  content  of  the  polygon  ABCDE, 
having  five  sides. 

Let  us  suppose  that  we  have  mea- 
sured the  diagonals  and  perpendicu- 
lars,  and    found    AC  =  36.21,   EC=: 


39.11,  B6=4,  J)d=7,26,  Aa- 


4.18,  required  the  area. 

Ans.  296.1292. 


PROBLEM  VI. 

To  find  the  area  of  a  long  and  irregular  figure,  bounded  on 
one  side  by  a  right  line. 

Rule. — 1.  At  the  extremities  of  the  right  line  measure  the  per- 
pendicular breadths  of  the  figure,  and  do  the  some  at  several 
intermediate  points,-  at  equal  distances  from  each  other. 
2.  Add  together  the  intermediate  breadths  and  haf  the  sum  of 
the  extreme  ones :  then  multiply  this  sum  by  one  of  the  equal 
parts  of  the  base  line :  the  j^roduct  will  be  the  required  area, 
very  nearly. 

Let  AEert  be  an  irregular  figure,  hav- 
ing for  its  base  the  right  line  AE.  At 
the  points  A,  B,  C,  D,  and  E,  equally 
distant  from  each  other,  erect  the  per- 
pendiculars Aa,  B6,  Cc,  Ddy  Ee,  to  the 


MENSURATION  OF  SURrACES.  27D 

base  line  AE,  and  designate  them  respectively  by  the  letters 
a,  h,  c,  6?,  and  e. 

Then,  tlie  area  of  the  trapezoid  ABZ?a= -'x  AB, 

the  area  of  the  trapezoid  BCc6=— I— xBC, 

c-{-d 
the  area  of  the  trapezoid  CDdc  — x  CD, 

d-\  e 
and  the  area  of  the  trapezoid  DEe<i= xDE  ; 

hence,  their  sum,  or  the  area  of  the  whole  figure,  is  equal  to 
/«  +  6     h-\-c     c-\-d     d+e\ 

since  AB,  BC,  &c.  are  equal  to  each  other.     But  this  sum  is 
also  equal  to 

(l.  +  b  +  c+d+l-)xAB, 
\2  2/ 

which  corresponds  with  the  enunciation  of  the  rule. 

1.  The  breadths  of  an  irregular  figure  at  five  equidistant 
places  being  8.2.  7.4,  9.2,  10.2,  and  8.6,  and  the  length  of  the 
base  40,  required  the  area. 


8.2  4)40 

8.6  — 


2(16.8 


10  one  of  the  equal  parts. 


8.4  mean  of  the  extremes. 

7.4  35.2    suni. 

9.2  10 

10.2  

352= area. 

35.2  sum.  


2.  The  length  of  an  irregular  figure  being  84,  and  the 
breadths  at  six  equidistant  places  17.4,  20.6,  14.2,  16.5,  20.1. 
and  24,4;  what  is  the  area?  Ans.  1550.64. 

PROBLEM  VII. 

To  find  the  area  of  a  regular  polygon. 

Rule  I. — Multiply  half  the  perimeter  of  the  pohjgon  hy  the 
apotheni,  or  perpendicular  let  fall  from  the  centre  on  one  of 
the  sides,  and  the  product  will  be  the  area  required  (Book  V, 
Prop.  IX.). 


280 


MENSURATION  OF  SURFACES. 


Re.ma.rk  I. — The  following  is  the  manner  of  detcrmmmg 
the  perpendicular  when  only  one  side  and  the  number  of  sides 
of  tiie  regular  polygon  are  known : — 

First,  divide  3G0  degrees  by  the  number  of  sides  of  the  poly- 
gon, and  the  (juotient  will  be  the  angle  at  the  centre  ;  that  is, 
the  angle  subtended  by  one  of  the  ecjual  sides.  Divide  this 
angle  by  2,  and  half  the  angle  at  the  centre  will  then  be  known. 

Now,  the  line  drawn  from  the  centre  to  an  angle  of  the 
polygon,  the  perpendicular  let  fall  on  one  of  the  equal  sides, 
and  half  this  side,  form  a  right-angled  triangle,  in  which  there 
are  knowii,  the  base,  which  is  half  the  equal  side  of  the  poly- 
gon, and  the  angle  at  the  vertex.  Hence,  the  perpendicular 
can  be  determined. 


1.  To  find  the  area  of  a  regular  hexa- 
gon, whose  sides  are  20  feet  each. 


G)3G0° 

eO°=:ACB,the  angle  at  the  centre. 
30°=ACD,  half  the  angle  at  the  centre 


Also,  CAD=90°— ACD=60°;  and  AD  =  ]0. 

Then,  as  sin  ACD  .  .  .  30%  ar.  comp 0.301030 

:    sin  CAD  ...  00^ 9.93753J 

•:   AD 10     1.000000 

:   CD  .  .  .  17.3205 '.  .  .  1.238561 

Perimeter  =120,  and  half  the  perimeter  =60. 
Then,  60  X  17.3205  =  1039.23,  the  area. 

2.  What  is  the  area  of  an  octagon  whose  side  is  20  ? 

Ans.  1931.36886. 

Remakk  TI. — The  area  of  a  regular  polygon  of  any  number 
of  sides  is  easily  calculated  by  the  above  rule.  Let  the  area? 
of  the  regular  polygons  whose  sides  are  unity,  or  1,  be  calcu- 
lated and  arranged  in  the  following 


MENSURATION  OF  SURFACEvS.  281 

TABIE. 

nics.  Sides.  Areas. 


Triangle    .     . 

3 

0.4330127 

S(|iiare      .     . 

.       4 

1.0000000 

Pentagon  .     . 

5 

1.7204774 

Hexagon    .     .     . 

.       6 

2.59807G2 

Heptagon  .     . 

.       7 

3.G339124 

Octagon     .     . 

.       8 

4.8284271 

Nonagon    ,     . 

.       9 

0.1818242 

Decagon    .     . 

.     10 

7.0942088 

Undecagon     . 

.     11 

9.3G50399 

Dodecagon     .     . 

.     12 

11.1901524 

Now,  since  the  areas  of  similar  polygons  ai^e  to  each  other 
as  tlie  scjuares  of  their  homologous  sides  (Book  IV.  Prop. 
XXVII.),  we  shall  have 

1-  :  tabular  area  : :  any  side  squared  :  area. 

Or,  to  find  the  area  of  any  regular  polygon,  we  have 

Rl'le  II. — 1.  Square  the  side  of  the  polygon. 

2.  Then  multiply  that  square  hy  the  tabular  area  set  opposite 
the  polygon  of  the  same  number  of  sides,  and  the  producl  will 
be  the  required  area. 

1.  What  is  the  area  of  a  regular  hexagon  whose  side  is  20? 

20-  =  400,      tabular  area  =2.5980702. 
Hence,  2.5980762x400=1039.2304800,  as  before. 

2.  To  find  the  area  of  a  pentagon  whose  side  is  25. 

Ans.   1075.298375. 

3.  To  find  the  area  of  a  decagon  whose  side  is  20. 

Ans.  3077.G8352. 

PROBLEM  Vlir. 

To  find  the  circumference  of  a  cir<:le  when  the  diameter  is 
given,  or  the  diameter  when  the  circumference  is  given. 

\\v\.T.. — Multiply  the  diameter  by  3.14 IG,  and  the  product  will 
he  the  circumference ;  or,  divide  the  circumference  by  3.1410, 
and  the  quotient  will  be  the  diameter. 

It  is  shown  (Book  V.  Prop.  XIV.),  that  the  circumference 

of  a  circle  whose  diameter  is  1,  is  3.141592G,  or  3.1410.     But 

since  the  circumferences  of  circles  are  to  each  other  as  their 

radii  or  diameters,  we  have,  by  calling  the  diameter  of  tlie 

second  circle  d, 

1   :  d  ::  3.141G  :  circumference, 

or,  cZx  3.1410=  circumference. 

TT             1                 J    circumference 
Hence,  also.  a= 

3.1410 

Aa2 


282      MENSURATION  OF  SURFACES. 

1.  What  is  the  circumference  of  a  circle  whose  diameter 
is  25  ?  .4715.  78.54. 

2.  If  the  diameter  of  the  earth  is  7921  miles,  what  is  the 
circumference?  Ans.  24884.()136. 

3.  Wliat  is  the  diameter  of  a  circle  whose  circumference  is 
11652.1904?  Ans.  37.09. 

4.  What  is  the  diameter  of  a  circle  whose  circumference  is 
6850?  '  Ans.  2180.41. 

PR^OBLEM  IX 

To  find  the  length  of  an  arc  of  a  circle  containing  any  numbei 
of  degrees. 

Rule. — Multiply  the  numher  of  degrees  in  the  given  arc  hy 
0.0087266,  and  the  product  hy  the  diameter  of  the  circle. 

Since  the  circumference  of  a  circle  whose  diameter  is  1,  is 
3.1416,  it  follows,  that  if  3.1416  bo  divided  by  360  degrees, 
the  quotient  will  be  the  length  of  an  arc  of  1  degree :  that  is, 

^•^'^^"=0.0087266=:  arc  of  one  degree  to  the  diameter  1. 

360 
This  being  multiplied  by  the  number  of  degrees  in  an  arc,  the 
product  will  be  the  length  of  that  arc  in  the  circle  whose  diam- 
eter is  1  ;  and  this  product  being  then  multiplied  by  the  diam- 
eter, will  give  the  length  of  the  arc  for  any  diameter  whatever. 

Remark. — When  the  arc  contains  degrees  and  minutes,  re- 
duce the  minutes  to  the  decimal  of  a  degree,  which  is  done  by 
dividing  them  by  60. 

1.  To  find  the  length  of  an  arc  of  30  degrees,  the  diameter 
being  18  feet.  Ans,  4.712364. 

2.  To  find  the  length  of  an  arc  of  12°  10',  or  12^°,  the  diam- 
eter being  20  feet.  Ans.  2.123472. 

3.  What  is  the  length  of  an  arc  of  10°  15',  or  10^°,  in  a  cir- 
cle whose  diameter  is  68  ?  Ans.  6.082396. 

PROBLEM  X. 
To  find  the  area  of  a  circle. 

Rule  I. — Multiply  the  circumference  hy  half  the  radius  (Book 

V.  Prop.  XII.). 
Rule  II. — Multiply  the  square  of  the  radius  hy  S.14l^  (Book 

V.  Prop.  XII.  Cor.  2). 

1.  To  find  the  area  of  a  circle  whose  diameter  is  10  and 
circumference  31.416.  Ans.  78.54. 


MENSURATION  OF  SURFACES.  283 

2.  Find  thp  area  of  a  circle  whose  diameter  is  7  and  cir- 
cunriference  21.9912.  Ans.  38.4846. 

3.  How  many  square  yards  in  a  circle  whose  diameter  is 
3^-  feet?  Ans.   1.069016. 

4.  What  is  the  area  of  a  circle  whose  circumference  is  12 
feet?  Ans.  11.4595. 

PROBLEM  XI. 
To  find  the  area  of  the  sector  of  a  circle. 

Rule  T. — Multiply  the  arc  of  the  sector  hy  half  the  radius  (Book 
V.  Prop.  XII.  Cor.  l). 

Rule  II. — Compute  the  area  of  the  whole  circle:  then  say^  as 
360  decrees  is  to  the  degrees  in  the  arc  of  the  sector,  so  is  the 
area  of  the  whole  circle  to  the  area  of  the  sector. 

1.  To  find  the  area  of  a  circular  sector  whose  arc  contains 
18  degrees,  the  diameter  of  the  circle  being  3  feet. 

Ans.  0.35343. 

2.  To  find  the  area  of  a  sector  whose  arc  is  20  feet,  the 
radius  being  10.  Ans.  100. 

3.  Required  the  area  of  a  sector  whose  arc  is  147°  29',  and 
radius  25  feet.         *  Ans.  804.3980. 

PROBLEM  XIL 
To  find  the  area  of  a  segment  of  a  circle. 

Rule. — 1.  Find  the  area  of  the  sector  having  the  same  arc,  hy 
the  last  problem. 

2.  Find  the  area  of  the  triangle  formed  hy  the  chord  of  the 
segment  and  the  two  radii  of  the  sector. 

3.  Then  add  these  two  together  for  the  answer  when  the  seg- 
ment is  greater  than  a  semicircle,  and  subtract' them  when  it 
is  less. 

1.  To  find  the  area  of  the  segment  p   • 

ACB,  its  chord  AB  being  12,  and  the  ^ 

radius  EA,  10  feet.  /^ 

AsEA      lOar.  comp.    .  .  9.000000  -^At" 

:  AD        6 0.778151  [      ^^^> 

::  sinD  90^ 10.000000 


sin  AED  30°  52' =  36.87  9.778151 
2 


73.74= the  degrees  in  the  arc  ACB. 


jiH«>. 


184  MENSURATION  OF  SURFACES. 

Then,  0.008726G  x  73.74  x  20  =  12.87  =  arc  ACB,  nearly. 

5  " 


G4.35  =  areaEACB. 


Again,      VEA'^— AD"  =  V 100—36=  n/G4=8=:ED; 
and  Gx8=:48  =  the  area  of  tlie  triangle  EAB. 

Hence,  sect.  EACH— EAB  =  64.35— 48  =  iG.35  =  ACB. 

2.  Find  the  area  of  the  segment  whose  height  is  18,  the 
diameter  of  the  circle  being  50.  Ans.  G3G.4834. 

3.  Required  the  area  of  the  segment  whose  chord  is  IG,  the 
diameter  being  20.  Ans.  44.7G4. 

PROBLEM  XIII.  ♦ 

To  find  the  area  of  a  circular  ring:  that  is,  the  area  included 
between  the  circumferences  of  two  circl-es  which  have  a 
common  centre. 

Rule. — TaJie  the  difference  between  the  areas  of  the  two  circles. 
Or,  subtract  the  square  of  the  less  radius  from  the  square  of 
the  greater i  and  multiply  the  remainder  by  3. HI 6. 

For  the  area  of  the  larger  is R-t 

and  of  the  smaller r-'t 


Their  difference, or  the  area  of  the  ring,  is     (R- — y-j-r. 

1.  The  diameters  of  two  concentric  circles  being  10  and  G, 
required  the  area  of  the  ring  contained  between  their  circum- 
ferences. Ans.  50.2G5G. 

2.  What  is  the  area  of  the  ring  when  the  diameters  of  the 
circles  are  10  and  20?  Ans.  235.G2. 

PROBLEM  XIV. 

To  find  the  area  of  an  ellipse,  or  oval.* 

Rule. — Multiply  the  two  semi-axes  tdgether^  and  their  product 
by  3.1416. 


r 


1.    Required  the  area  of  an  ellipse 
whose  semi-axes  AE,  EC,  are  35  and  25. 
Ans.  2748.9. 


*  Although  this  rule,  and  the  one  for  (he  following  prollem,  cannot  be  de- 
monstrated without  the  aid  of  principles  not  yet  C(wiFi(lered,  ^till  it  was  tlion{jl»t 
host  to  insert  them,  as  they  coinplelo  tlic  rules  necessary  for  the  mensuration 
of  planes. 


MENSURATION  OF  SOLIDS.  285 

2.  Required  the  area  of  an  ellipse  whose  axes  are  24  and  18. 

Ans.  339.2928. 

PROBLEM  XV. 

To  find  the-  area  of  any  portion  of  a  parabola. 

Rule. — Multiply  the  base  by  the  perpendicular  height,  and  take 
two-thirds  of  the  product  for  the  required  area. 

C 

1.  To  find  the  area  of  the  parabola 
ACB,  the  base  AB  being  20  and  the  al- 
titude CD,  18. 

^715.  240. 

/  ^  A 

2.  Required  the  area  of  a  parabola,  the  base  being  20  and 
the  altitude  30.  Ans,  400. 


MENSURATION  OF  SOLIDS. 

The  mensuration  of  solids  is  divided  into  two  parts. 

1st.  The  mensuration  of  their  surfaces  ;  and, 

2dly.  The  mensuration  of  their  solidities. 

We  have  already  seen,  that  the  unit  of  measure  for  plane 
surfaces  is  a  square  whose  side  is  the  unit  of  length. 

A  curved  line  which  is  expressed  by  numbers  is  also  referred 
to  a  unit  of  length,  and  its  numerical  value  is  the  number  of 
times  which  the  line  contains  its  unit.  If,  then,  we  suppose  the 
linear  unit  to  be  reduced  to  a  right  line,  and  a  square  con- 
structed on  this  line,  this  square  will  be  the  unit  of  measure 
for  curved  surface^/ 

The  unit  of  solidity  is  a  cube,  the  face  r»f  which  is  equal  to 
the  superficial  unit  in  which  the  surface  of  the  solid  is  estimated, 
and  the  edge  is  equal  to  the  linear  unit  in  which  the  linear  di- 
mensions of  the  solid  are  expressed  (Book  VII.  Prop.  XIII. 
Sch.). 

The  following  is  a  table  of  solid  measures : — 


1728 

cubic  inches 

=     1  cubic  foot. 

27 

cubic  feet 

=     1  cubic  yard. 

44921 

cubic  feet 

=     1  cubic  rod. 

282 

cubic  inches 

=     1  ale  gallon. 

231 

cubic  inches 

=     1  wine  gallon 

2150.42 

cubic  inches 

=     1  bushel. 

280  MENSURATION  OF  SOLIDS. 


OF    POLYEDRONS,    OR    SURFACES    BOUNDED    BY    PLANES. 

PROBLEM  I. 

To  find  the  surface  of  a  right  prism. 

Rule. — Multiply  the  perimeter  of  the  base  hi/  the  altitude,  and 
the  product  will  be  the  convex  surface  (Book  VII.  Prop.  I.). 
To  this  add  the  area  of  the  two  bases,  when  the  entire  surface 
is  I'equired. 

1.  To  find  the  surface  of  a  cube,  the  length  of  each  side 
being  20  feet.  Ans.  2400  sq.ft. 

2.  To  find  the  whole  surface  of  a  triangular  prism,  whose 
base  is  an  equilateral  triangle,  having  each  of  its  sides  equal 
to  18  inches,  and  altitude  20  feet.  Ans.  91.949. 

3.  What  must  be  paid  for  lining  a  rectangular  cistern  with 
lead  at  2d.  a  pound,  the  thickness  of  the  lead  being  such  as  to 
require  libs,  for  each  square  foot  of  surface  ;  the  inner  dimen- 
sions of  the  cistern  being  as  follows,  viz.  the  length  3  feet  2 
inches,  the  breadth  2  feet  8  inches,  and  the  depth  2  feet  C  inches  ? 

^     ,  Ans.  21.  3s.  lO'^d. 

PROBLEM  IL 

To  find  the  surface  of  a  regular  pyramid. 

Rule. — Multiply  the  perimeter  of  the  base  by  haf  the  slant 
height,  and  the  product  will  be  the  convex  surface  (Book  VII. 
Prop.  IV.) :  to  this  add  the  area  of  the  base,  when  the  entire 
surface  is  required. 

1.  To  find  the  convex  surface  of  a  regular  triangular  pyra- 
mid, the  slant  height  being  20  feet,  and  each  side  of  tlie  base 
3  feet.  Ans.  90  sq.ft. 

2.  What  is  the  entire  surface  of  a  regular  pyramid,  w'hose 
slant  height  is  15  feet,  and  the  base  a  pentagon,  of  which  each 
side  is  25  feet  ?  Ans.  2012.798. 

PROBLEM  III. 

To  find  the  convex  surface  of  the  frustum  of  a  regular 
pyramid. 

Rule. — Multiply  the  half  sum  of  the  perimeters  of  the  two 
bases  by  the  slant  hei<rht  of  the  fn/stum.  and  the  product  will 
be  the  convex  surface  (Book  VlL  Prop.  iV.  Cor.). 


MENSURATION  OF  SOLIDS.  287 

1.  How  many  square  feet  are  there  in  the  convex  surface  of 
he  frustum  of  a  square  pyramid,  whose  slant  height  is  10  feet, 

each  side  of  the  lower  base  3  feet  4  inclies,  and  each  side  of 
the  upper  base  2  feet  2  inches?  Ans.  110  sq.  ft. 

2.  Wiiat  is  the  convex  surface  of  the  frustum  of  an  hepta- 
gona^  pyramid  whose  slant  iieight  is  55  feet,  each  side  of  the 
lower  base  8  i'eet,  and  each  side  of  the  upper  base  4  leet  ? 

Ans.  2310  s^.  ft. 

PROBLEM  IV 

To  find  the  soli-dity  of  a  prism. 

Rule. — 1.  Find  the  area  of  the  base. 

2.  Muhip'y  the  area  of  the  base  by  the  altitude,  and  the  pro- 
duct will  be  the  soliditij  of  the  prism  (Book  VII.  Prop.  XIV.). 

1.  What  is  the  solid  content  of  a  cube  whose  side  is  24 
inches?  '  Ans.  13824. 

2.  How  many  cubic  feet  in  a  block  of  marble,  of  which  the 
length  is  3  feet  2  inches,  breadth  2  feet  8  inches,  and  height  or 
thickness  2  feet  6  inches?  Ans.  21^. 

3.  How  many  gallons  of  water,  ale  measure,  will  a  cistern 
contain,  whose  dimensions  are  the  same  as  in  the  last  example  ? 

Ans.  129if 

4.  Required  the  solidity  of  a  triangular  prism,  whose  height 
is  10  feet,  and  the  three  sides  of  its  triangular  base  3, 4,  and  5 
feet.  Ans.  60. 

PROBLEM  V, 

To  find  the  solidity  of  a  pyramid. 

Rule. — Multiply  the  area  of  the  base  by  one-third  of  the  alti- 
tude, and  the  product  will  be  the  solidity  (Book  VII.  Prop. 
XVII.). 

1.  Required  the  solidity  of  a  square  pyramid,  each  side  of 
its  base  being  30,  and  the  altitude  25.  Ans.  7500. 

.   2.  To  find  the  solidity  of  a  triangular  pyramid,  whose  alti- 
tude is  30,  and  each  side  of  the  base  3  feet.    Ans.  38.9711. 

3.  To  find  the  solidity  of  a  triangular  pyramid,  its  altitude 
being  14  feet  6  inches,  and  the  three  sides  of  its  base  5,  6,  and 
7  feet.  Ans.  71.0352. 

4.  What  is  the  solidity  of  a  pentagonal  pyramid,  its  altitude 
being  12  feet,  and  each  side  of  its  base  2  feet  ? 

Ans.  27.5276. 

5.  What  is  the  solidity  of  an  hexagonal  pyramid,  whose  alti- 
tude is  6.4  feet,  and  each  side  of  its  base  6  inches  ? 

Ans.  1.38564. 


2S8  MENSURATION  OI-  SOLIDS. 

PROBLEM  VI. 
To  find  the  solidity  of  the  frustum  of  a  pyramid. 

Rule. — Add  together  the  areas  of  the  two  bases  of  the  frustum 
and  a  mean  proportional  between  them,  and  then  multiply  the 
sum  by  one-third  of  the  altitude  (Book  VII.  Prop.  XVllI.). 

1.  To  find  the  number  of  solid  feet  in  a  piece  of  timber, 
whose  bases  are  squares,  each  side  of  the  lower  base  being  15 
inches,  and  each  side  of  the  upper  base  6  inches,  the  altitude 
being  24  feet.  Ans.  19.5. 

2.  Required  the  solidity  of  a  pentagonal  frustum,  whose  alti- 
tude is  5  feet,  each  side  of  the  lower  base  18  inches,  and  each 
side  of  the  upper  base  6  inches.  Ans.  9.31925. 


Definitions. 

1.  A  wedge  is  a  solid  bounded  by  five  c?  H 
pian^ps :  viz.  a  rectangle  ABCD,  called 
the  base  of  the  wedge  ;  two  trapezoids 
ABHG,  DCHG,  which  are  called  the 
sides  of  the  wedge,  and  which  intersect 
each  other  in  the  edge  GH  ;  and  the  two 
triangles  GDA,  HCB,  which  are  called 
the  ends  of  tlie  wedge. 

When  AB,  the  length  of  the  base,  is  equal  to  GH,  the  trape- 
zoids ABHG,  DCHG,  become  parallelograms,  and  the  wedge 
is  then  one-half  the  parallelopipedon  described  on  the  base 
ABCD,  and  having  the  same  altitude  with  the  wedge. 

The  altitude  of  the  wedge  is  the  perpendicular  let  fall  from 
any  point  of  the  line  GH,  on  the  base  ABCD. 

2.  A  rectangular  prismoid  is  a  solid  resembling  the  frustum 
of  a  quadrangular  pyramid.  The  upper  and  lower  bases  are 
rectangles,  having  their  corresponding  sides  parallel,  and  the 
convex  surface  is  made  up  of  four  trapezoids.  The  altitude  of 
the  prismoid  is  the  perpendicular  distance  between  its  bases. 


PROBLEM  VIL 
To  find  the  solidity  of  a  wedge. 

Rule. — To  twice  the  length  of  the  base  add  the  length  of  the 
edge.  Multiply  this  sum  by  the  breadth  of  the  base,  and  then 
by  the  altitude  of  the  wedge,  and  take  one-sixth  of  the  product 
for  the  solidity. 


MENSURATION  OF  SOLIDS. 


289 


the 


length 


Let  L=AB, 

the  base. 

/=GH,  the  length 
the  edge. 

b=BC,  the  breadth  of 
the  base. 

/i=PG,  the  altitude 
the  wedge. 

Then,    L— Z=AB— GH  = 
AM. 

Suppose  AB,  the  length  of  the  base,  to  be  equal  to  GH,  the 
length  of  the  edge,  the  solidity  will  then  be  equal  to  half  the 
parallelopipedon  having  the  same  base  and  the  same  altitude 
(Book  VII.  Prop.  Yll.y.  Hence,  the  solidity  will  be  equal 
to  iblh  (Book  VII.  Prop.  XIV.).' 

If  the  length  of  the  base  is  greater  than  that  of  the  edge, 
let  a  section  MNG  be  made  parallel  to  the  end  BCH.  The 
wedge  will  then  be  divided  into  the  triangular  prism  BCH-M, 
and  the  quadrangular  pyramid  G-AMND. 

The  solidity  of  the  prism  =^bhl,  the  solidity  of  the  pyramid 
-=^bh(h—l);  and  their  sum,  ibhl+^bh(L—r)=}bhSl+ibh2h 
— J-fe/i2/=jM(2L  +  /). 

If  the  length  of  the  base  is  less  than  the  length  of  the  edge, 
the  solidity  of  the  wedge  will  be  equal  to  the  difference  be- 
tween the  prism  and  pyramid,  and  we  shall  have  for  the  solid- 
ity of  the  wedge, 

yblil—^bh{l—L)  =}bh2l—l  bh2l+l  bh2L=ibh{2L  +  /). 

1.  If  the  base  of  a  wedge  is  40  by  20  feet,  the  edge  35  feet, 
and  the  altitude  10  feet,  what  is  the  solidity  ? 

Ans.  3833.33. 

2.  The  base  of  a  wedge  being  18  feet  by  9,  the  edge  20 
feet,  and  the  altitude  6  feet,  what  is  the  solidity  ? 

Ans,  504. 


PROBLEM  VIII. 

To  find  the  solidity  of  a  rectangular  prismoid. 

Rule. — Add  together  the  areas  of  the  two  bases  and  four  times 
the  area  of  a  parallel  section  at  equal  distances  from  the 
bases :  then  multiply  th$  sum  by  one-sixth  of  the  altitude. 

Bb 


290  MENSURATION  OF  SOLIDS. 

Let  L  and  B  be  the  length  and 
breadth  of  the  lower  base,  /  and  h  the 
length  and  breadth  of  the  upper  base, 
M  and  m  the  length  and  breadth  of  the 
section  equidistant  from  the  bases,  and 
h  the  altitude  of  the  prismoid. 

Through  the  diagonal  edges  L  and 
t  let  a  plane  be  passed,  and  it  will  di- 
vide the  prismoid   into  two  wedges, 
having  for  bases,  the  bases  of  the  prismoid,  and  for  edges  the 
lines  L  and  l'=L  J 

The  solidity  of  these  wedges,  and  consequently  of  the  pris-    ' 
moid,  is 

But  since  M  is  equally  distant  from  L  and  /,  we  have 
2M=L  +  /,         and  2m=:B  +  b', 
hence,        4Mm=(L+/)  x  (B  +  6)=BL  +  BZ+6L+R 

Substituting  4Mm  for  its  value  in  the  preceding  equation, 
and  we  have  for  the  solidity 

i/i(BL+M+4Mm). 

Remark. — This  rule  may  be  applied  to  any  prismoid  what- 
ever. For,  whatever  be  the  form  of  the  bases,  there  may  be 
inscribed  in  each  the  same  number  of  rectangles,  and  the  num- 
ber of  these  rectangles  may  be  made  so  great  that  their  sum 
in  each  base  will  differ  from  that  base,  by  less  than  any  assign- 
able quantity.  Now,  if  on  these  rectangles,  rectangular  pris- 
moids  be  constructed,  their  sum  will  differ  from  the  given  pris- 
moid by  less  than  any  assignable  quantity.  Hence  the  rule  is 
general. 

1.  One  of  the  bases  of  a  rectangular  prismoid  is  25  feet  by 
20,  the  other  15  feet  by  10,  and  the  altitude  12  feet ;  required 
the  solidity.  Ans.  3700. 

2.  What  is  the  solidity  of  a  stick  of  hewn  timber,  whose 
endg  are  30  inches  by  27,  and  24  inches  by  18,  its  length  being 
24  feet?  Ans.  102  feet. 


OP   THE    MEASURES    OP   THE    THREE    ROUND   BODIES, 

PROBLEM  IX. 

To  find  the  surface  of  a  cylinder. 

Rule. — Multiply  the  circumference  of  the  base  by  the  altitude, 
and  the  product  will  be  the  convex  surface  (Book  VIII.  Prop. 
I.).  To  this  add  the  areas  of  the  two  bases ,  when  the  entire 
surface  is  required. 


MENSURx\TION  OF  SOLIDS.  291 

I      - 

1.  What  is  the  convex  surface  of  a  cylinder,  the  diameter 

of  whose  base  is  20,  and  whose  altitude  is  50  ? 

Ans.  3141.6. 

S       2.  Required  the  entire  surface  of  a  cylinder,  whose  altitude 

is  20  feet,  and  the  diameter  of  its  base  2  feet. 

^715.  131.9472. 


■  PROBLEM  X. 

'  To  find  the  convex  surface  of  a  cone. 

Rule. — Multiply  the  circumference  of  the  base  by  half  the  side 
\  (Book  VIII.  Prop.  III.)  :  to  which  add  the  area  of  the  base, 
I       when  the  entire  surface  is  required. 

f       1.  Required  the  convex  surface  of  a  cone,  whose  side  is  50 
feet,  and  the  diameter  of  its  base  8^  feet.  Ans.  667.59. 

2.  Required  the  entire  surface  of  a  cone,  whose  side  is  36 
and  the  diameter  of  its  base  18  feet.  Ans.  1272.348. 

PROBLEM  XI. 

To  find  the  surface  of  the  frustum  of  a  cone. 

Rule. — Multiply  the  side  of  the  frustum  by  half  the  sum  of  the 
circumferences  of  the  two  bases,  for  the  convex  surface  (Book 
VIII.  Prop.  IV.) :  to  which  add  the  areas  of  the  two  bases, 
when  the  entire  surface  is  required. 

1.  To  find  the  convex  surface  of  the  frustum  of  a  cone,  the 
side  of  the  frustum  being  12^  feet,  and  the  circumferences  of 
the  bases  8.4  feet  and  6  feet.  Ans.  90. 

2.  To  find  the  entire  surface  of  the  frustum  of  a  cone,  the 
side  bemg  16  feet,  and  the  radii  of  the  bases  3  feet  and  2  feet. 

Ans.  292.1688. 

PROBIiEM  XII. 

To  find  the  solidity  of  a  cylinder. 

Rule. — Multiply  the  area  of  the  base  by  the  altitude  (Book  VIII. 
Prop.  II.). 

1.  Required  the  solidity  of  a  cylinder  whose  altitude  is  12 
feet,  and  the  diameter  of  its  base  15  feet.        Ans.  2120.58. 

2.  Required  the  solidity  of  a  cylinder  whose  altitude  is  20 
feet,  and  the  circumference  of  whose  base  is  5  feet  6  inches. 

Arts.  48.144. 


292  MENSURATION  OF  SOLIDS. 

PROBLEM  XIII. 

To  find  the  solidity  of  a  cone. 

Rule. — Multiply  the  area  of  the  base  by  the  altitude^  and  take 
one-third  of  the  product  (Book  VIII.  Prop.  V.). 

1.  Required  the  soHdity  of  a  cone  whose  aUitude  is  27  feet, 
and  the  diameter  of  the  base  10  feet.  Ans.  706.86. 

2.  Required  the  sohdity  of  a  cone  whose  altitude  is  10^ feet, 
and  the  circumference  of  its  base  9  feet.       Ans.  22.56. 

PROBLEM  XIV. 

To  find  the  solidity  of  the  frustum  of  a  cone. 

Rule. — Add  together  the  areas  of  the  two  bases  and  a  mean 
proportional  between  them,  and  then  multiply  the  sum  by  one- 
third  of  the  altitude  (Book  VIII.  Prop.  VI.). 

1.  To  find  the  solidity  of  the  frustum  of  a  cone,  the  altitude 
being  18,  the  diameter  of  the  lower  base  8,  and  that  of  the 
upper  base  4.  Ans.  527.7888. 

2.  What  is  the  solidity  of  the  frustum  of  a  cone,  the  altitude 
being  25,  the  circumference  of  the  lower  base  20,  and  that  of 
the  upper  base  10?  Ans.  464.216. 

3.  If  a  cask,  which  is  composed  of  two  equal  conic  frustums 
joined  together  at  their  larger  bases,  have  its  bung  diameter  28 
inches,  the  head  diameter  20  inches,  and  the  length  40  inches 
how  many  gallons  of  wine  will  it  contain,  there  being  231  cubic 
inches  in  a  gallon  ?  Ans.  79.0613. 

PROBLEM.  XV. 

To  find  the  surface  of  a  sphere. 

-iix"   Rule  I. — Multiply  the  circumference  of  a  great  circle  by  the 
diameter  (Book  VIII.  Prop.  X.). 

Rule  II. — Multiply  the  square  of  the  diameter^  or  four  times 
the  square  of  the  radius,  by  3.1416  (Book  VIII.  Prop.  X. 
Cor.). 

1.  Required  the  surface  of  a  sphere  whose  diameter  is  7. 

Ans.  153.9384. 

2.  Required  the  surface  of  a  sphere  whose  diameter  is  24 
inches.  Ans.  1809.5616  in. 

3.  Required  the  area  of  the  surface  of  the  earth,  its  diam- 
eter being  7921  miles.  Ans.   197111024  sq.  miles. 

4.  What  is  the  surface  of  a  sphere,  the  circumference  of  its 
great  circle  being  78.54?  Ans.  1963.5. 


MENSURATION  OF  SOLIDS.  293 

PROBLEM  XVI. 
To  find  the  surface  of  a  spherical  zone. 

Rule. — Multiply  the  altitude  of  the  zone  by  the  circumference 
of  a  great  circle  of  the  sphere,  and  the  product  will  he  the 
surface  (Book  Vlll.  Prop.  X.  Sch.  1). 

1.  The  diameter  of  a  sphere  being  42  inches,  what  is  the 
convex  surface  of  a  zone  whose  altitude  is  9  inches  ? 

Ans.  USl[.5248sq.in, 

2.  If  the  diameter  of  a  sphere  is  12^  feet,  what  will  be  the 
surface  of  a  zone  whose  altitude  is  2  feet? 

Ans,  78.54  sq,  ft, 

PROBLEM  XVII. 
To  find  the  solidity  of  a  sphere. 

Rule  I. — Multiply  the  surface  by  one-third  of  the  radius  (Book 
VIII.  Prop.  XIV.). 

Rule  II. — Cube  the  diameter,  and  multiply  the  number  thus 
found  by  ^rt ;  that  is,  by  0.5236  (Book  VIII.  Prop.  XIV. 
Sch.  3). 

1.  What  is  the  solidity  of  a  sphere  whose  diameter  is  12? 

Ans.  904.7808. 

2.  What  is  the  solidity  of  the  earth,  if  the  mean  diameter 
be  taken  equal  to  7918.7  miles  ?  Ans,  259992792083. 

PROBLEM  XVIII. 
To  find  the  solidity  of  a  spherical  segment. 

Rule. — Find  the  areas  of  the  two  bases,  and  multiply  their  sum 
by  half  the  height  of  the  segment ;  to  this  product  add  the- 
solidity  of  a  sphere  whose  diameter  is  equal  to  the  height  of 
the  seginent  (Book  VIII.  Prop.  XVII.). 

Remark. — ^When  the  segment  has  but  one  base,  the  other  is 
to  be  considered  equal  to  0  (Book  VIII.  Def.  14). 

1.  What  is  the  solidity  of  a  spherical  segment,  the  diameter 
of  the  sphere  being  40,  and  the  distances  from  the  centre  to  the 
bases,  16  and  10.  Ans.  4297.7088. 

2.  What  is  the  solidity  of  a  spherical  segment  with  one  base, 
the  diameter  of  the  sphere  being  8,  and  the  altitude  of  the 
seffment  2  feet?  Ans,  41.888 

Bb2 


294  MENSURATION  OF  SOLIDS. 

3.  What  is  the  solidity  of  a  spherical  segment  with  one  base, 
the  diameter  of  the  sphere  being  20,  and  the  altitude  of  the 
segment  0  feet  ?  Ans.  1781.2872. 

PROBLEM  XIX. 

To  find  the  surface  of  a  spherical  triangle. 

Rule. — 1.  Compute  the  surface  of  the  sphere  on  which  the  trian- 
gle is  formed,  and  divide  it  by  8  ;  the  quotient  will  be  the  sur- 
face of  the  tri-rectangular  triangle, 

2.  Add  the  three  angles  together ;  from  their  sum  subtract 
180"^,  and  divide  the  remainder  by  90^ :  then  multiply  the  tri- 
rectangular  triangle  by  this  quotient,  and  the  product  will  be 
the  surface  of  the  triangle  (Book  IX.  Prop.  XX.). 

1.  Required  the  surface  of  a  triangle  described  on  a  sphere, 
whose  diameter  is  30  feet,  the  angles  being  140"^,  92°,  and  68°. 

Ans.  471.24  sq.ft. 

2.  Required  the  surface  of  a  triangle  described  on  a  sphere 
of  20  feet  diameter,  the  angles  being  120°  each. 

Ans.  Sl4:.l6  sq.ft. 

PROBLEM  XX. 
To  find  the  surface  of  a  spherical  polygon. 

Rule. — 1.  Find  the  tri-rectangular  triangle,  as  before. 

2.  From  the  sum  of  all  the  angles  take  the  product  of  two 
right  angles  by  the  number  of  sides  less  two.  Divide  the  re- 
mainder by  90°,  and.  multiply  the  tri-rectangular  triangle  by 
the  quotient :  the  product  will  be  the  surface  of  the  polygon 
(Book  IX.  Prop.  XXL). 

1.  What  is  the  surface  of  a  polygon  of  seven  sides,  de- 
scribed on  a  sphere  whose  diameter  is  17  feet,  the  sum  of  the 
angles  being  1080°  ?  Ans.  226.98. 

2.  What  is  the  surface  of  a  regular  polygon  of  eight  sides, 
described  on  a  sphere  whose  diameter  is  30,  each  angle  of  the 
polygon  being  140°  ?  ^715.157.08. 


OF  THE    REGULAR    POLYEDRONS. 

In  determining  the  solidities  of  the  regular  polyedrons,  it 
becomes  necessary  to  know,  for  each  of  them,  the  angle  con- 
tained between  any  two  of  the  adjacent  faces.  The  determi- 
nation of  this  angle  involves  the  following  property  of  a  regu- 
lar polygon,  viz. — 


MENSURATION  OF  SOLIDS. 


295 


Half  the  diagonal  wliich  joins  the  extremities  of  two  adjacent 
sides  of  a  regular  polygon,  is  equal  to  the  side  of  the  polygon 
multiplied  hy  the  cosine  of  the  angle  which  is  obtained  by  di- 
viding 360°  by  twice  the  number  of  sides :  the  radius  being 
equal  to  unity. 


Let  ABODE  be  any  regular  poly- 
gon. Draw  the  diagonal  AC,  and  from 
the  centre  F  draw  FG,  perpendicular 
to  AB.  Draw  also  AF,  FB ;  the  lat- 
ter will  be  perpendicular  to  the  diag- 
onal AC,  and  will  bisect  it  at  H  (Book 
in.  Prop.  VI.  Sch.). 

Let  the  number  of  sides  of  the  poly- 
gon be  designated  by  n :  then, 

AFB =??2!,        and  AFG  =.  CAB 


360'= 


n  2n 

But  in  the  right-angled  triangle  ABH,  we  have 

AH=AB  cos  A-AB  cos  —-   (Trig.  Th.  I.  Cor.) 

271 


Remark  1 . — ^When  the  polygon  in  question  is  the  equilateral 
triangle,  the  diagonal  becomes  a  side,  and  consequently  half 
the  diagonal  becomes  half  a  side  of  the  triangle. 

The  perpendicular  BH=AB  sin  frl  (Trig. 


Remark  8. 
Th.  I.  Cor.). 


2n 


To  determine  the  angle  included  between  the  two  adjacent 
faces  of  either  of  the  regular  polyedrons,  let  us  suppose  a  plane 
to  be  passed  perpendicular  to  the  axis  of  a  solid  angle,  and 
through  the  vertices  of  the  solid  angles  which  lie  adjacent. 
This  plane  will  intersect  the  convex  surface  of  the  polyedron 
in  a  regular  polygon  ;  the  number  of  sides  of  this  polygon  will 
be  equal  to  the  number  of  planes  which  meet  at  the  vertex  of 
either  of  the  solid  angles,  and  each  side  will  be  a  diagonal  of 
one  of  the  equal  faces  of  the  polyedron. 

Let  D  be  the  vertex  of  a  solid  angle, 
CD  the  intersection  of  two  adjacent  faces, 
and  ABC  the  section  made  in  the  convex 
surface  of  the  polyedron  by  a  plane  per- 
pendicular to  the  axis  through  D. 

Through  AB  let  a  plane  be  drawn  per- 
pendicular to  CD,  produced  if  necessary, 
and  suppose  AE,  BE,  to  be  the  lines  in 


2*J6 


MENSURATION  OF  SOLIDS. 


which  this  plane  intersects  the  adjacent 
faces.  Then  will  AEB  be  the  angle  in- 
cluded between  the  adjacent  faces,  and 
FEB  will  be  half  that  angle,  which  we 
will  represent  by  ^A. 

Then,  if  we  represent  by  n  the  num- 
ber of  faces  which  meet  at  the  vertex  of  j^ 
the  solid  angle,  and  by  m  the  number  of 
sides  of  each  face,  we   shall  have,  from  what  has 
been  shown, 

300° 


BF=:BC  cos  ??2!, 


and  EB=BC  sin 


2n 


But 


hence, 


BF 


2m 


— =sin  FEB = sin  ^A,  to  the  radius  of  unity ; 
£B 


cos 


sin  ^A=. 


360° 
~2n' 


sm 


360° 


42' 


This  formula  gives,  for  the  plane  angle  formed  by  every  two 
adjacent  faces  of  the 

Tetraedron 70°  31' 

Hexaedron .  90° 

Octaedron   .     .     .     .     .     .     .     .  109°  28' 

Dodecaedron 116° 

Icosaedron 138° 


18" 
33'  54" 
ir  23' 


Having  thus  found  the  angle  included  between  the  adjacent 
faces,  we  can  easily  calculate  the  perpendicular  let  fall  from 
the  centre  of  the  polyedron  on  one  of  its  faces,  when  the  faces 
themselves  are  known. 

The  following  table  shows  the  solidities  and  surfaces  of  the 
regular  ()olyedrons,  when  the  edges  are  equal  to  1. 

A  TABLE  OP  THE  REGULAR  POLYEDRONS  WHOSE  EDGES  ARE  1. 


Names. 

No.  of  Faces. 

Tetraedron  . 

.  .  .  .     4  .  .  . 

Hexaedron   . 

...     6  .  .  . 

Octaedron.  . 

...     8  .  .  . 

Dodecaedron 

.  .  .  .  12  .  .  . 

Icosaedron   . 

.  .  .  20  .  .  . 

Surface.  Solidity. 

1.7320508  ....  0.1178513 

6.0000000  .  .  .  :  1.0000000 

3.4641016  ....  0.4714045 

20.0457288  ....  7.6631189 

8.6602540  ....  2.1816950 


MENSURATION  OF  SOLIDS  297 

PROBLEM  XXI. 

To  find  the  solidity  of  a  regular  polyedron. 

Rule  I. — Multiply  the  surface  hy  one-third  of  the  perpendicular 
let  fall  from  the  centre  on  one  of  the  faces ^  and  the  product 
will  be  the  solidity. 

Rule-  II. — Multiply  the  cube  of  one  of  the  edges  by  the  solidity 
of  a  similar  polyedron,  whose  edge  is  1. 

The  first  rule  results  from  the  division  of  the  polyedron  into 
as  many  equal  pyramids  as  it  has  faces.  The  second  is  proved 
by  considering  that  tw^o  regular  polyedrons  having  the  same 
number  of  faces  may  be  divided  into  an  equal  number  of  simi- 
lar pyramids,  and  that  the  sum  of  the  pyramids  which  make 
up  one  of  the  polyedrons  will  be  to  the  sum  of  the  pyramids 
which  make  up  the  other  polyedron,  as  a  pyramid  of  the  first 
sum  to  a  pyramid  of  the  second  (Book  II. 'Prop.  X.) ;  that  is, 
as  the  cubes  of  their  homologous  edges  (Book  VII.  Prop.  XX.) ; 
that  is,  as  the  cubes  of  the  edges  of  the  polyedron. 

1.  What  is  the  solidity  of  a  tetraedron  whose  edge  is  15? 

lAns.  397.75. 

2.  What  is  the  solidity  of  a  hexaedron  whose  edge  is  12? 

Ans,  1728. 

3.  What  is  the  solidity  of  a  octaedron  whose  edge  is  20  ? 

Ans.  3771.236. 

4.  What  is  the  solidity  of  a  dodecaedron  whose  edge  is  25  ? 

Ans.  119736.2328. 

5.  What  is  the  solidity  of  an  icosaedron  whose  side  is  20  ? 

Ans,  17453.56. 


V  ^ 


A   TABLE 


OF 


LOGARITHMS    OF    JTUMBERS 


FROM     1     TO     10,000. 


N. 

IjOC. 

N. 

Lofi. 

N. 

l^f'K. 

N. 

Loff. 

1 

0.000000 

26 

1.414973 

51 

1.707570 

76 

1.880814 

2 

0.301030 

27 

1.431364 

52 

1.716003 

77 

1.886491 

3 

0.477121 

28 

1.447158 

53 

1.724276 

78 

1.892095 

4 

0.602060 

29 

1.462398 

54 

1.732394 

79 

1.897627 

5 

0.698970 

30 

1.477121 

55 

1.740.363 

80 

1.903090 

6 

0.778151 

31 

1.491362 

56 

1.748188 

81 

1 . 908485 

7 

0.845098 

32 

1.505150 

57 

1.7.55875 

82 

1.913814 

8 

0.903090 

33 

1.518514 

58 

1.763428 

83 

1.919078 

9 

0.954243 

34 

1.531479 

59 

1.770852 

84 

1.924279 

If) 

1.000000 

35 

1.544068 

60 

1.778151 

85 

1.929419 

li 

1.041393 

36 

1.556303 

61 

1.785330 

86 

1.934498 

12 

1.079181 

37 

1.. 568202 

62 

1.792392 

87 

1.939519 

13 

1.113943 

38 

1.579784 

63 

1.799341 

88 

1.944483 

14 

1.146128 

39 

1.591065 

64 

1.806180 

89 

1.949390 

15 

1.176091 

40 

1.602060 

G5 

1.812913 

90 

1.954243 

16 

1.204120 

41 

1.612784 

66 

1.819544 

91 

1.959041 

17 

1.230449 

42 

1.623249 

67 

1.826075 

92 

1.963788 

18 

1.255273 

43 

1.633468 

08 

1.832509 

93 

1.968483 

19 

1.278754 

44 

1.643453 

69 

1.838849 

94 

1.973128 

"zO 

1.3010.30 

45 

1.653213 

70 

1.845098 

95 

1.977724 

21 

1.322219 

46 

1.662758 

71 

1.851258 

96 

1.982271 

22 

1.342423 

47 

1.672098 

72 

1.8.57333 

97 

1.986772 

23 

1.361728 

48 

1.681241' 

73 

1.863323 

98 

1.991226 

24 

1.380211 

49 

1.690196 

74 

1.869232 

99 

1.995635 

25 

1.397940 

50 

1.698970 

75 

1.875061 

100 

2.000000 

N.B.  In  the  following  table,  in  the  last  nine  columns  of  each 
page,  Avhere  the  first  or  leading  figures  change  from  9's  to  O's, 
points  or  dots  are  introduced  instead  of  the  O's  through  the  rest 
Df  the  line,  to  catch  the  eye,  and  to  indicate  that  from  thence 
the  annexed  first  two  figures  of  the  Logarithm  in  the  second 
column  stand  in  the  next  lower  line. 


1 


A  TABLE  OF  LOGARITHMS  FROM   1  -£0  10,000. 


IN.  1   0   1  1  1  2  1  3  i  4  i  5'  i  6  1  7  1  8  1  9  1  D.  I 

100 

OOOOOOi  0434 

0868 

1301 

1734 

2166 

2598 1  3029 

3461 

3891 

432 

101 

4321 

4751 

5181 

5609 

6038 

6466 

6S94 

7321 

7748 

8174 

428 

102 

8G00 

9026 

9451 

9876 

.300 

.724 

1147 

1570 

1993 

2416 

424 

103 

012837 

3259 

3680 

4100 

4531 

4940 

5360 

5779 

6197 

6616 

419 

104 

7033 

7451 

7868 

8284 

8700 

9116 

9532 

9947 

.361 

.775 

416 

105 

021189 

1603 

2016 

2428 

2841 

3252 

3664 

4075 

4486 

4896 

412 

106 

5306 

5715 

,6125 

6533 

6942 

7350 

7757 

8164 

8571 

8978 

408 

107 

9384 

9789 

.195 

.600 

1004 

1408 

1812 

2216 

2619 

3021 

404 

108 

033424 

3826 

4227 

4628 

5029 

5430 

5830 

6230 

6629 

7028 

400 

109 
110 

7426 

7825 
1787 

8223 
2182 

8620 
2576 

9017 
2969 

9414 
3362 

9811 
3755 

.207 

4148 

.602 
4540 

.998 
4932 

396 
393 

041393 

111 

5323 

5714 

6105 

6495 

6885 

7275 

7664 

8053 

8442 

8830 

389 

112 

9218 

9606 

9993 

.380 

.766 

1 1.53 

1638 

1924 

2309 

2694 

386 

113 

053078 

3463 

3846 

4230 

4613 

4996 

5378 

5700 

6142 

6524 

382 

114 

6905 

7286 

7666 

8046 

8426 

8805 

9185 

9563 

9942 

.320 

379 

115 

060698 

1075 

1452 

1829 

2206 

2582 

2958 

3333 

3709 

4083 

376 

116 

4458 

4832 

5206 

5580 

5953 

6326 

G699 

7071 

7443 

7815 

372 

117 

8186 

8557 

8928 

9298 

9668 

..38 

.407 

.776 

1145 

1514 

369 

118 

071882 

2250 

2617 

2985 

3352 

3718 

4085 

4451 

4816 

5182 

366 

119 
120 

5547 

5912 
9543 

6276 
9904 

6640 
.266 

7004 
.626 

7368 
.987 

7731 
13"47 

8094 
1707 

8457 
2067 

8819 
2426 

363 
3§0 

079181 

121 

082785 

3144 

3503 

3861 

4219 

4576 

4934 

5291 

5647 

6004 

357 

122 

6360 

6716 

7071 

7426 

7781 

8136 

8490 

8845 

9198 

9552 

355 

123 

9905 

.258 

.611 

.963 

1315 

1667 

2018 

2370 

2721 

3071 

351 

124 

093422 

3772 

4122 

4471 

4820 

5169 

6518 

5866 

6215 

6562 

349 

125 

6910 

7257 

7G04 

7951 

8298 

8644 

8990 

9335 

9681 

..26 

346. 

126 

100371 

0715 

1059 

1403 

1747 

2091 

2434 

2777 

3119 

3462 

343 

127 

3804 

4146 

4487 

4828 

5169 

5510 

5851 

6191 

6531 

6871 

340 

128 

7210 

7549 

7888 

8227 

8565 

8903 

9241 

9579 

9916 

.253 

338 

129 
130 

110590 

0926 

4277 

1263 
4611 

1599 
4944 

1934 

2270 

2605 
5943 

2940 
6276 

3275 
6608 

3809 
6940 

335 
333 

113943 

5278 

5611 

131 

7271 

7603 

7934 

8265 

8595 

8926 

9256 

9586 

9915 

.245 

330 

132 

120574 

0903 

1231 

1560 

188S 

2216 

2544 

2871 

3198 

3525 

328 

133 

3852 

4178 

4504 

4830 

5156 

5481 

5806 

6131 

6456 

6781 

325 

134 

7105 

74291 

7753 

8076 

8399 

8722 

9045 

9368 

9690 

..12 

823 

135 

130334 

06551 

0977 

1298 

1619 

1039 

2260 

2580 

2900 

3219 

321 

136 

3539 

3858 

4177 

4496 

4814 

5133 

5451 

5769 

6086 

6403 

318 

137 

6721 

7037 

7354 

7671 

7987 

8303 

8618  8934 

9249 

9564 

315 

138 

9879 

.194 

.508 

.822 

1136 

1450 

1763 

2076 

2389 

2702 

314 

139 

140 

143015 

3327 
6438 

3639 

6748 

3951 

7058 

4263 
7337 

4574 
7676 

4885 
7985 

5196 

8294 

5507 
8603 

5818 
8911 

311 
309 

146128 

141 

9219 

9527 

9835 

.142 

.449 

.756 

1063 

1370 

1676 

1982 

307 

142 

152288 

2594 

2900 

3205 

3510 

3815 

4120 

4424 

4728 

5032 

305 

143 

5336 

6640. 

5943 

6216 

6549 

6852 

7154 

7457 

7759 

8061 

303 

144 

8362 

8664 

8965 

9266 

9567 

9868 

.168 

.469 

.769 

1068 

.301 

U6 

161368 

1667 

1987 

2266 

2564 

2863 

3161 

3460 

3758 

4055 

299 

146 

4353 

4650 

4947 

5244 

5541 

5838 

6134 

6430 

6726 

7022 

297 

147 

7317 

7613 

7908 

8203 

8497 

8792 

9086 

9.380 

9674 

9968 

295 

148 

170262 

0555 

0848 

1141 

1434 

1726 

2019  2311 

2603 

2895 

293 

149 
150 

3186 

3478 
6381 

3769 
6670 

4060 
6959 

4351 

7248 

4541 
7536 

4932 

7825 

5222 
8113 

5512 
8401 

5802 
8689 

291 
289 

176091 

151 

8977 

9264 

9552 

9839 

.126 

3270 

.699 

.985 

1272 

1558 

287 

152 

181844 

2129 

2415 

2700 

2985 

3555  3839 

4123 

4407 

285 

153 

4691 

4975 

5259 

5542 

5825 

6108 

6391  6674 

6956 

7239 

283 

154 

7521 

7803 

8084 

8366 

8647 

8928 

9209 

9490 

9771 

..51 

281 

155 

190332 

0612 

0892 

1171 

1451 

1730 

2010 

2289 

2567 

2846 

279 

156 

3125 

3403 

3681 

3959 

4237 

4514 

4792 

5069 

5346 

5623 

278 

157 

5899 

6176 

6453 

6729 

7005 

7281 

7556 

7832 

8107 

8382 

276 

158 

8657 

8932 

9206 

9481 

9755 

..29 

.303 

.577 

.8.50 

1124 

274 

159 

201397 

1670 

1943 

2216 

2488 

2761 

3033 

3305 

3577 

3848 

272 

N.  1   0   1  1  1  2  1  3  1  4  1  5  1  6  1  7  1  8  1  9  1  D.  1 

A  TABLE  OP  LOGAKITIIMS  FROM   1   TO  10,000. 


M.  1   0   1  I   I  2  1  3  1  4  1  5  1  G  1  7  1  8  1  9  1  n.  1 

~lW 

204120 

4391 

4663  4934 

5204 

5475 

5740 

6016;  6286 

6556 

271 

IGl 

6826 

7096 

7305  7634 

7904 

8173 

8441 

87101  8979 

9247 

269 

102 

9515 

9783 

..51  .319 

.586 

.853 

1121 

13S8| 1654 

1921 

26? 

163 

212 188 

2454!  2720i  2986 

3252 

3518 

3T83 

4049J4314 

4579 i  266  | 

104 

4844 

5109 

5373 

5638 

5932 

6100 

0430 

6694' 6957 

7221 

264 

165 

7484 

7747 

8010 

8273 

8538 

8798 

9060 

9323  9535 

9846 

262 

lOfi 

220108 

0370 

0631 

0892 

1153 

1414 

1075 

1936  2196 

2456 

201 

107 

2716 

2976 

3236 

3490 

3755 

4015 

4274 

4533  4792 

.5051 

259 

168 

6309 

5568 

5826 

0084 

0342 

6600'  6858 

71151  7372;  7630 

258 

169 

7887 

8144 

8400 

8057 

8913 

9170 

9420 

9082  9933 

.193 

256 

170 

23U449 

07041 0960 

1215 

1470 

1724 

1979 

2234  2488 

2742 

2.54 

171 

2996 

3250 

3504 

3757 

4011 

4264 

4517 

4770  .5023 

5276 

253 

172 

5528 

5781 

6033 

6285 

0537 

6789 

7041 

7292!  7544 

7/95 

252 

173 

8046 

8297 

8548 

8799 

9049 

9299 

9550 

9800 

1  ...50 

.300 

250 

174 

240549 

0799 

1048 

1297 

1546 

1795 

2044 

2293 

2541 

2790 

249 

175 

303S 

3280 

3534 

3782 

4030 

4277 

4525 

4772 

.5019 

5266 

248 

176 

5513 

5759 1  6006 

6252 

6499 

674-5 

0991 

7237 

7482 

7728 

240 

177 

7973 

8219  8464 

8709 

8954 

9198 

9443 

9687 

9932 

.176 

245 

178 

250420 

0664' 0908 

1151 

1395 

1638 

1881 

2125 

2368 

2610 

243 

179 

2853 

30'96i  33381  3580 

3822 

4004 

4300 

4548 

4790 

5031 

242 

180 

255273 

5514  5755 

5996 

6237 

6477 

6718 

6958 

7198 

7439 

241 

181 

7079 

7918  8158 

8398 

8637 

8877 

9116 

9355 

9594 

9833 

239 

182 

260071 

0310  0548 

0787 

1025 

1263 

1501 

1739 

1976 

2214 

238 

183 

2451 

2688  2925 

3162 

3399|  .3636 

3873 

4109 

4346 

4582 

237 

184 

4818 

5054  5290 

5525 

5761 

5995 

6232 

6407 

6702 

6937 

235 

185 

7172 

7406  7641  7875 

8110 

8,344 

8578 

8812 

9046 

9279 

234 

186 

9513 

9746  9980 

.213 

.446 

.079 

.912 

1144 

1377 

1609 

233 

187 

271842 

2074 

2306 

2538 

2770 

3001 

3233 

3464 

3696 

3927 

232 

188 

4158 

4389 

4620 

4850 

5081 

5311 

5542 

5772 

6002 

6232 

230 

189 
190 

6462 

6692 

8982 

6921 
92'l  f 

7151 
9439 

7380  7609 

78381  8067i  8296 

8525 
.806 

229 
228 

278754 

9667 

98951  .1231  ..351 

.578 

191 

281033 

1261 

1488 

1715 

1942 

2109  2396 

2622 

2849 

3075 

227 

192 

3301 

3527 

3753 

3979 

4205 

4431  4656 

4882 

5107 

5332 

220 

193 

5557 

5782 

6007 

6232 

6456 

6681  6905 

7130 

7354 

7578 

225 

194 

7802 

8026 

8249 

8473 

8696 

8920 

9143 

9366 

9589 

9812 

223 

195 

290035 

0257 

0480 

0702 

0925 

1147 

1369 

1.591 

1813 

2034 

222 

196 

2256 

2478 

269i» 

2920 

3141 

3363 

3584 

3804 

4025 

4246 

221 

197 

4466 

4687 

4907 

5127 

5347 

5567 

5787  600-;  1  G226I 

6446 

220 

198 

6065 

6884 

7104 

7323 

7542(7761 

7979  8198 

8416 

8635 

219 

199 

200 

8853 
301030 

9071 
1247 

9289 
1464 

9507 
1681 

9725 

1898 

9943 

.161 
2331 

.378 
2547 

.595 

2764 

.813 
2980 

218 
217 

2114 

201 

3196 

3412 

3628 

3844 

4059 

4275 

4491 

4706 

4921 

5136 

216 

202 

5351 

5566 

5781 

5996 

6211 

6.425 
^64 

6639 

6854 

7068 

7282 

215 

203 

7496 

7710 

7924 

8137 

8351 

8778 

8991 

9204 

9417 

213 

204 

9030 

9843 

..56 

.268 

.481 

.693 

.906 

1118 

1330 

1.542 

212 

205 

311754 

1986 

2177 

2389 

2600 

2812 

3023 

3234 

3445 

3656 

211 

200 

3807 

4078 

4289 

4499 

4710 

4920 

5130 

5340 

.5551 

5760 

210 

207 

5970 

6180 

6390 

6599 

6809 

7018 

7227 

7436 

7646 

7854 

209 

208 

8063 

8272 

8481 

8089 

8898 

9100 

9314 

9522 

9730 

9938 

208 

209 
210 

320146 

0354 

2426 

0562 
2633 

0709 
2839 

0977 
3046 

1184 
3252 

1391 
3458 

1598 
3665 

1805 
3871 

2012 
4077 

207 
206 

322219 

211 

4282 

4488 

4694 

4899 

5105 

5310 

5516 

.5721 

5926 

6131 

205 

212 

6336 

6541 

6745 

0950 

7155 

7359 

7563 

7767 

7972 

8176 

204 

213 

8380 

8583 

8787 

8991 

9194 

9398  9001 

9805 

...8 

.211 

203 

214 

330414 

0017 

0819 

1022 

1225 

1427 

1630 

1832 

2034 

2236 

202 

215 

2438 

2040 

2842 

3044 

3246 

3447 

3049 

3850 

4051 

4253  202 

216 

4454 

4655 

4856 

5057 

.5257 

.5458 

5658 

.5859 

60.59 

6260  201 

217 

6460 

66601 6860 

7060 

7260 

7459 

7659 

7858 

8058 

8257  200 

218 

8456 

86561  885-5 

9054 

9253 

9451 

9650 

9349  ..47! 

.246  199 

219 

340444  0642' 0841 1  1039 

1237' 1435' 1632'  1830' 2028' 

2225  198 

N  i   0   1  I  1  2  1  3  1  4  1  5  1  6  1  7  1  8  i  9  j  D.l 

15* 


CO 


A  TABLE  OF  LOGARITHMS  FROM  1  TO  lO.OOG. 


jN. 

0   1  1  1  2  I  3  1  4  1  5  1  6  1  7  1  8  1  9  '(  D.  1 

230 

342423, 2620 

2817!  3014  3212,  3409,  3oi)6  3802 

399i4|41<vfii  197 

221 

4392 

4589 

4785 

4981 

5178 

5374 

5570 

6766 

69621 81571  196 

222 

6353 

6549 

6744 

6939 

7135 

9083 

7330 

7525 

7720 

7915]  8110!  195 

223 

8305 

8500 

8694 

8889 

9278 

9472 

9666 

98G0i  ..64|  194 

224 

350248 

0442 

0636  0829 

105i3 

1216 

1410 

1603 

1796! 1989| 193 

225 

2183 

2375 

2568 1 2761 

2954 

31-17 

3339 

3532 

3724139161  193 

226 

4108 

4301 

4493 

4685 

4876 

5068 

6260 

5452 

661315834!  192 

227 

6026 

6217 

6408 

6599 

6790 

6981 

7172 

7363 

7554  77'14i  191 

228 

7935 

8125 

8316 

8506 

8696 

8888 

9076 

9266 

^456!  96461  190 

229 
23J 

9835 

..25 
1917 

.215 
2105 

.404 
2294 

.593 

2482 

.783 
2671 

.972 
2859 

1161 
3048 

1350  1539:  189 
3238  3124;  188 

361728 

231 

3612 

3800 

3988 

4176 

4363 

4551 

4739 

4928 

5113  630 1|  188 

232 

5488 

5675 

5862 

6049 

6236 

6423 

6610 

0796 

8983j7169|  187 

233 

7356 

7542 

7729 

7915 

8101 

8287 

8473 

8659 

8845  9030 i  186 

234 

9216 

9401 

9587 

9772 

9958 

.143 

.328 

.513 

.698  .883' 185 

235 

371068 

1253 
3098 

1437 

1622 

1806 

1991 

2175 

2360 

2544  2728' 184 

236 

2912 

3280 

3464 

3G47 

3331 

4015 

4198 

4332  456-3;  184 

237 

4748 

4932 

5115 

6298 

5481 

5064 

584G 

6029 

6212 

6394:'  183 

238 

6577 

6759 

6942 

7124 

7306 

7488 

7670 

7852 

8034 

82l6i  182 

239 

240 

8398 

8580 

8761 
0573 

8943 
0754 

9124 
0934 

9306 
1115 

9487 
1298 

9668 
1476 

9849 
1656 

ll30 

1837 

181 
181 

380211 

0392 

241 

2017 

2197 

2377 

2557 

2737 

2917 

3097 

.3277 

3456 

3836 

180 

242 

3815 

3995 

4174 

4353 

4533 

4712 

4891 

5070 

5249 

5428 

179 

243 

5606 

5785 

5964 

6142 

6321 

6499 

0677 

6856 

7034 

7212 

178 

244 

7390 

7568 

7746 

7923 

8101 

8279 

8456 

8634 

8811 

89891 178 

245 

9166 

9343 

9520 

9698 

9875 

..51 

.228 

.405 

.582 

.759! 177 

246 

390935 

1112 

1288 

1464 

1641 

1817 

1993 

2189 

2345 

2521 

176 

247 

2697 

2873 

3048 

3224 

3400 

3575 

3751 

3926 

4101 

4277 

176 

248 

4452 

4627 

4802 

4977 

5152 

5326 

5501 

5876 

5850 

6025 

175 

249 

6199 

6374 

6548 

6722 

6896 

7071 

7245 

7419 

7592 

7766 

174 

250 

397940 

8114 

8237 

8481 

8634 

8808 

8981 

9154 

9328 

9501 

173 

251 

9674 

9847 

..20 

.192 

.365 

.538 

.711 

.883 

1056 

12281 173 

252 

401401 

1573 

1745 

1917 

2089 

2261 

2433 

2605 

2777 

2949! 172 

253 

3121 

3292 

3464 

3635 

3807 

3978 

414S 

4320 

4492;  48831  171  | 

254 

4834 

5005 

5176 

6346 

55.7 

5688 

5858 

6029 

6199 

6370 

171 

255 

6540 

6710 

6881 

7051 

7221 

7391 

7661 

7731 

7901 

8070 

170 

256 

8240 

8410 

8579 

8749 

8918 

9087 

9257 

9426 

9595 

9764 

169 

257 

9933 

.102 

.271 

.440 

.609 

.777 

.946 

1114 

1283 

1451 

189 

258 

411620 

1788 

1956 

2124 

2293 

2461 

2629 

2796 

2964 

3i32| 168 

259 
260 

3300 

3467 
5140 

3635 
5307 

3803 

5474 

3970 
6641 

4J37 
6808 

4305 

5974 

4472 
6141 

4639 
6308 

4808 1 167 
6474!  167 

414973 

281 

6641 

6807 

6973 

7139 

7306 

7472 

7638 

7804 

7970181351  166 

262 

8301 

8467 

8633 

8798 

8964 

9129 

9295 

9160 

9625!  979 1|  165 

263 

9956 

.121 

.286 

.451 

•616 

.781 

.945 

1110 

1275 

14391  165 

264 

421604 

17C8 

1933  2097 

■2261 

2426 

2590 

2754 

2918 

3082i  164 

9.65 

•3246 

3410 

3574  3737 

3901 

4065 

4228 

4392 

4555 

4718!  164 

266 

4882 

5045 

5208 

5371 

6534 

5097 

5860 

6023 

6186 

6349 

163 

267 

6511 

6674 

6S36 

6999 

716] 

7324 

7488 

7648 

7811 

7973 

162 

268 

8135 

8297 

8459 

8621 

8783 

8944 

9106 

9268 

9429 

9591 

162 

269 

9752 

9914 

..75 

.236 

.398 

.559 

.720 

.881 

1042 

1203 

161 

270 

431364 

1525 

1685 

1846 

2007 

2167 

2328 

2488 

J>349 

2809!  181  1 

271 

2969 

3130 

3290 

3450 

3610 

3770 

3930 

4090 

4249144091  160  f 

272 

4569 
6163 

4729 

4883 

5048 

5207 

5307 

6626 

5685 

6844 i  6004 

1 59 

273 

6322 

6481 

6640 

6798 

6957 

7116 

7275 

7433' 7592 

159 

274 

7751 

7904 

8087 

8226 

8384 

8542 

8701 

8869 

^017  9175 
.694  .752 

158 

275 

9333 

9491 

9648 

9806 

9964 

.122 

.279 

.437 

1.08 

276 

440909 

1066 

1224 

1381 

1538 

1695 

1852 

2009 

216612323 

157 

277 

2480 

'?537 

2793 

2950 

3106 

3263 

3419 

3576 

373213889;  1571 

278 

4045 

4201 

4357 

4513 

4069 

4825 

4981 

6137 

529315449  156 

279 

6604 

5760 

59151  60711  62261  6382  6537  6692  6848' 70031  155 | 

N. 

1   0 

i  1 

i  2 

1  3 

1  4 

5 

6 

7 

8 

9 

D.I 

A  TABLE  OF  LOGARITHMS  FROM   1   TO   10,000. 


iN. 

0   1  1  i  2  1  3  1  4  1  5  1  6  1  7  !  8  1  9  1  D.  1 

280 

447158 

7313 

7468 

7623 

7778 

7933 

8088 

8242 

8397 
9941 

8552 

155 

281 

8700 

8861 

9015 

9170 

9324 

9478 

9633 

9787 

..95 

154 

282 

450249 

0403 

0557 

0711 

0865 

1018 

1172 

1326 

1479 

1633 

154 

283 

1786 

1940 

2093 

2247 

2400 

2553 

2706 

2859 

3012 

3165 

153 

284 

3318 

3471 

3624 

3777 

3930 

4082 

4235 

4387 

4540 

4692 

1.53 

285 

4845 

4997 

5150 

5302 

5454 

5600 

5758 

5910 

6062 

6214 

152 

286 

6366 

6518 

6670 

6821 

6973 

7125 

7276 

7428 

7579 

7731 

152 

287 

7882 

8033 

8184 

8336 

8487 

8638 

8789 

8940 

9091 

9242 

151 

288 

9392 

9543 

9694 

9845 

9995 

.146 

.296 

.447 

.597 

.748 

151 

289 

460S9S 

1048 

119.8 

1348 

1499 

1649 

1799 

1948 

2098 

2248 

150 

290 

462398 

2548 

2697 

2847 

2997 

3146 

3296 

3445 

3594 

3744 

150 

291 

3893 

4042 

4191 

4340 

4490 

4639 

4788 

4936 

6085 

5234 

149 

292 

5383 

5532 

5680 

5829 

5977 

6126 

6274 

6423 

6571 

6719 

149 

293 

6868 

7016 

7164 

7312 

7460 

7608 

7756 

7904 

8052 

8200 

148 

294 

8347 

8495 

8643 

8790 

8938 

9085 

9233 

9380 

9527 

9675 

148 

295 

9822 

9969 

.116 

.263 

.410 

.55* 

.704 

.851 

.998 

1145 

147 

296 

471292 

1438 

1585 

1732 

1878 

2025 

2171 

2318 

2464 

2610 

146 

297 

2756 

2903 

3049 

3195 

3341 

3487 

3633 

3779 

3926 

4071 

146 

298 

4216 

4362 

4508 

4653 

4799 

4944 

5090 

5235 

5381 

5526 

146 

299 

5671 

5816 

5962 

6107 

6252 

6397 

6542 

6687 

6832 

6976 

145 

300 

477121 

7266 

7411 

7555 

7700 

7844 

7989 

8133 

8278 

8422 

146 

301 

8566 

8711 

8855 

899  EJ 

0143 

9287 

9431 

9575 

9719 

9863 

144 

302 

480007 

0151 

0294 

0438 

0582 

0725 

0869 

1012 

1156 

1299 

144 

303 

1443 

1586 

1729 

1872 

2016 

2159 

2302 

2445 

2588 

2731 

143 

304 

2874 

3016 

3159 

3302 

3445 

3587 

3730 

3872 

4015 

4157 

143 

305 

4300 

4442 

4585 

4727 

4869 

5011 

5153 

5295 

5437 

5579 

142 

306 

5721 

5863 

6006 

6147 

6289 

6430 

6572 

6714 

6856 

6997 

.42 

307 

7138 

7280 

7421 

7563 

7704 

7845 

7986 

8127 

8269 

8410 

141 

308 

8551 

8692 

8833 

8974 

9114 

9255 

9396 

9537 

9677 

9818 

141 

309 
310 

9958 

..99 
1502 

.239 
1642 

.380 

.520 
1922 

.661 
2062 

.601 

2201 

.941 
2341 

1081 
2481 

1222 
2621 

140 
140 

491362 

1782 

311 

2760 

2900 

3040 

3179 

3319 

3458 

3597 

3737 

3876 

4015 

1.39 

312 

4155 

4294 

4433 

4572 

4711 

4850 

4989 

5128 

6267 

5406 

139 

313 

5544 

5683 

5822 

5960 

6099 

6238 

6376 

6515 

6653 

6791 

139 

314 

6930 

7068 

7206 

7344 

7483 

7621 

7759 

7897 

8035 

8173 

138 

315 

8311 

8448 

8586 

8724 

8862 

8999 

9137 

9275 

9412 

9550 

138 

316 

0687 

9824 

9962 

..99 

.236 

.3/4 

.511 

.648 

.785 

.922 

137 

317 

501059 

1196 

1333 

1470]  1607 

1744 

1880 

2017 

2154 

2291 

137 

318 

2427 

2564 

2700 

2837!  2973 

3109 

3246 

3382 

3518 

3655 

136 

319 

3791 

.3927 

4063 

4199 

4335 

4471 

4607 

4743 

4878 

5014 

136 

320 

505150 

5286 

5421 

5557 

5693 

5828 

5964 

6099 

6234 

6370 

136 

321 

6505 

6640 

6776 

6911 

7046 

7181 

7316 

7451 

7586 

7721 

135 

322 

7856 

7991 

8126 

8260 

8395 

8530 

8664 

8799 

8934 

9068 

135 

323 

9203 

9337 

9471 

9606 

9740 

9874 

...9 

.143 

.277 

.411 

134 

324 

510545 

0679 

0813 

0947 

1081 

1215 

1349 

1482 

1016 

1750 

.134 

325 

1883 

2017 

2151 

2284 

2418 

2551 

2684 

2818 

2951 

3084 

133 

326 

3218 

3351 

3484 

3617 

3750 

3883 

4016 

4149 

4282 

4414 

133 

327 

454S 

4681 

4813 

4946 

5079 

5211 

5344 

5476 

5609 

5741 

133 

328 

5874 

6006 

6139 

6271 16403 

6535 

6668 

6800 

6932 

7064 

132 

329 

7196 

7328 

7460 

7592 

7724 

7855 

7987 

8119 

8251 

8382 

132 

330 

518514 

8646 

8777 

8909 

9040 

9171 

9303 

9434 

95C6 

9697 

131 

331 

9828 

9959 

..90 

.221 

.353 

.484 

.615 

.745 

.876 

1007 

131 

332 

521138 

1269 

1400 

1530 

1661 

1792 

1922 

2053 

2183 

2314 

131 

333 

2444 

2575 

2705 

2835 

2966 

3096 

3226 

3356 

3486 

36  J  6 

130 

334 

3746 

3876 

4006 

4136 

4266 

4396 

4526 

4656 

4785 

4915 

130 

335 

5045 

5174 

5304 

5434 

5563 

5693 

5822 

5951 

6081 

0210 

129 

338 

6339 

6469 

6598 

6727| 6856 

6985 

7114 

7243 

7372 

7.501 

129 

337 

7630 

7759 

7888 

8016.1  8145 

8274 

8402 

8531 

8660 

8788'  129 

338 

8917 

9045 

9174 

9302i  9430 

9559 

9687 

9815 

9943 

..72| 128 

339 

5302001  03281  0456' 0584i  0712 

084010968' 1096 

1223 

13511  128 

0   1  1   1  2  1  3  !  4  i  5  I  6  1  7  !  8  1  9  1  i).  i 

^'^ww 


0 

A  TABLE  OF  LOGARITHMS  F1103I  1 

TO  10,000 

;^-  i   0   1  1  1  2  1  3  1  4  1  5  1  6  i  7  1  S  1  9  1  I).  1 

340 

5314791  1607 

1734 

1  1862 

1990 

2117 

1 22451  2372 

2500 

^i)27 

128 

341 

2754 

2882 

3009 

31.36 

3264 

3391 

I3518 

3645 

3772 

3899 

127 

342 

4026 

4153 

4280 

4407 

4534 

4661 

4787 

4914 

5041 

5167 

127 

343 

5294 

5421 

5547 

5674 

5800 

5927 

6053 

6180 

6306 

6432 

126 

344 

6558 

6685 

6811 

6937 

7063 

7189 

7315 

7441 

7567 

7693 

126 

345 

7819 

7945 

8071 

8197 

8322 

8448 

8574 

8699 

8825 

8951 

126 

34G 

9076 

9202 

9327 

9452 

9578 

9703 

9829 

9954 

..79 

.204 

125 

347 

540329 

0455  0580 

0705 

0830 

0955 

1080 

1205 

1330 

14.54 

125 

348 

1579 

1704 

1829 

1953 

2078 

2203 

2327 

2452 

2576 

2701 

125' 

349 

2825 

2950 

3074 

3199 

3323 

3447 

3571 

3696 

38ro 

3944 

124 

350 

544068 

4192 

4316 

4440 

4564 

4688 

4812 

4936 

5060 

5183 

124 

351 

5307 

5431 

5555 

5078 

5802 

6925 

6049 

6172 

629^ 

6419 

124 

352 

6543 

6666 

6789 

6913 

7036 

7159 

7282 

7405 

7529 

7652 

123 

353 

7775 

7898 

8021 

8144 

8267 

8389 

8512 

8635 

87.58 

8881 

123 

354 

9003 

9126 

9249 

9371 

9494 

9616 

9739 

9861 

9984 

.106 

123 

355 

550228 

0351 

0473 

0595 

0717 

0840 

0962 

1084 

1206 

13281  122  1 

356 

1450 

1572 

1694 

1816 

19a8 

2060 

2181 

2303 

2425 

2547 

122 

357 

2068 

2790 

2911 

3033 

31.55 

3276 

3398 

3519 

3640 

3762 

121 

358 

3883 

4004 

4126 

4247 

4368 

4489 

4610 

4731 

4852 

4973 

121 

359 

360 

5094 

5215 

5336 

6544 

5457 
6664 

5578 
6785 

5699 
6905 

5820 
7026 

6940 
7146 

6061 
7267 

6182 

7387 

121 
120 

556303 

6423 

361 

7507 

7627 

7748 

7868 

7988 

8108 

8228 

8349 

8469 

8589 

120 

362 

8709 

8829 

8948 

9068 

9188 

9308 

9428 

9548 

9667 

9787 

120 

363 

9907 

..26 

.146 

.265 

.385 

.504 

.624 

.743 

.863 

.982 

1!9 

364 

561101 

1221 

1.340 

1459 

1578 

1698 

1817 

19.36 

2055 

2174 

119 

365 

2293 

2412 

2531 

2650 

2769 

2887 

3006 

3125 

3244 

3362 

119 

366 

3481 

3600 

3718 

.3837 

3955 

4074 

4192 

4311 

4429 

4548 

119 

367 

4666 

4784 

4903 

5021 

5139 

5257 

5376 

5494 

5612 

5730 

118 

368 

6848 

5966 

6084 

6202 

6320 

6437 

6555 

6673 

6791 

6909 

118 

369 

7026 

7144 

7262 

7379 

7497 

7614 

7732 

7849 

7967 

8084 

118 

370 

568202 

8319 

8436 

8554 

8671 

8788" 

8905 

902:^ 

9140! 

9257 

117 

371 

9374 

9491 

9608 

9725 

9812 

9959 

..76 

.193  .309i 

.426 

117 

372 

570543 

0660 

0776 

0893 

1010 

1126 

1243 

1359 

1476 

1592 

117 

373 

1709 

1825 

1942 

2058 

2174 

2291 

2407 

2523 

2639 

2755 

116 

374 

2872 

2988 

3104 

3220 

3336 

3452 

3568 

3684 

3800 

3915 

116 

375 

4031 

4147 

4263 

4379 

4494 

4610 

4726 

4841 

4957 

5072 

116 

376 

5188 

5303 

5419 

5534 

5650 

5765 

5880 

.5996 

6111 

6226 

115 

377 

6341 

6457 

6572 

6687 

6802 

6917 

7032 

7147 

7262 

7377 

115 

378 

7492 

7607 

7722 

7836 

7951 

8066 

8181 

8295 

8410 

8525 

115 

379 
380 

8639 
579784 

8754 
9898 

8868 
..12 

8983 
.126 

9097 
.241 

9212 
.355 

9320 
.469 

9441 

9555 

.697 

9669 
.811 

114 
114 

..583 

381 

580925 

1039 

1153 

1267 

1381 

1495 

1608 

1722 

1836 

1950 

114 

382 

2063 

2177 

2291 

2404 

2518 

2631 

2745 

2858 

2972 

3085 

114 

383 

3199 

3312 

3426 

3539 

3652 

3765 

3879 

3992 

4105 

4218 

113 

384 

4331 

4444 

4557 

4670 

4783 

4896 

6009 

5122 

5235 

5348 

113 

385 

5461 

5574 

5686 

5799 

.5912 

6024 

61.37 

6250 

6362 

0476 

113 

386 

6587 

6700 

6812 

6925 

7037 

7149 

7262 

7374 

7486 

7599 

112 

387 

7711 

7823 

7935 

8047 

8160 

8272 

83S4 

8496 

8608 

8720 

112 

388 

8832 

8944 

9056 

9167 

9279 

9391 

9503 

9615 

9726 

9838 

112 

3c9 

9950 

..61 

.173 

.284 

..S96 

.507 

.619 

.730 

.843 

.953 

112 

390 

591065 

1 176 

1287 

1399 

1510 

1621 

1732 

1843 

1935 

2066 

111 

391 

2177 

2288 

2399 

2510 

2621 

2732 

2843 

2954 

3064  3175 

111 

392 

3286 

3397 

3508 

3618 

3729 

38401 

3950 

4061 

4171  4282 

III 

393 

4393 

4503 

4014 

4724 

4834 

4945 

.5055 

5]  65 

.5276  53S6 

110 

394 

5496 

5606 

5717 

.5827 

5937 

6047 

61.57 

626. 

6377  6487 

no 

395 

6597 

6707 

6817 

6927 

7037 

7146! 

7256 

7366 

7476  7586 

110 

396 

7695 

7805 

7914 

8024 

8  KM 

8243  i 

8353 

8462 

8572i  8681 

110 

397 

8791 

8900 

9009 

9119 

922S 

9;w7 

9446 

95.56 

9665,  ^^774 

109 

3:)8 

9883 

9992 

.101 

.210 

.3191 

.428 

.537 

.646 

.755i  «64 

109 

399 

600973 

I0S2 

119! 

1299  14081 

1517 

1 625 

L731 

1843.  ly;.l 

109 

N.  1   0   1  1  1  2  1  3  i  4  1  5  i  6  '[  7  1  8  1  9  !  P.  1 

A  TABLE  OF  LOGARITHMS  FROM   1   TO  10.000. 


N.    I 


I     1     I     2 


4     !     5     I     6     I     7     i     8     I     9     I   D 


400 
401 
V)^ 
103 
404 
405 
406 
407 
408 
'100 
4l0 
'111 
412 
413 
414 
415 
416 
417 
418 
4_19 
420 
421 
422 
423 
424 
425 
420 
127 
428 
4vi9 
4S0 
431 
432 
433 
431 
435 
436 
437 
438 
439 


440 
441 
442 
443 
444 
445 
446 
447 
44  S 
.449 
450 
451 
452 
453 
454 
455 
456 
457 
45S 
45!) 

¥7 


602060 

4-2169 

1  2277 

2386 

3  I  'l>tBP*2f»3 

13361 

3469 

422^334 

4442 

4550 

5305 

5413 

5521 

6628 

6381 

6489 

6596 

6704 

7455 

7562 

'7669 

7777 

8526 

8633 

8740 

8847 

9594'  9701 

9808 

9914 

610660 

0767 

0873 

0979 

1723 

,  1829 

'  2890 

1936 
2996 

2042 
3102 

612i'84 

3842 

3947 

4053 

4159 

489? 

5003 

6108 

5213 

5950 

6055 

6160 

6265 

7000 

7105 

7210 

7315 

8048 

8153 

8257 

8362 

9093 

9198 

9302 

9106 

620136 

0240 

0344 

0448 

1176 

12S0 

1384 

1488 

2214 

2318 

2421 
3456 

2525 
3559 

623249 

3353 

4282 

4385 

4488 

4591 

5312 

5415 

5518 

6621 

6340 

6443 

6546 

6648 

7366 

7468 

7571 

7673 

8389 

8491 

8593 

8695 

9410 

9512 

9613 

9716 

630428 

0530 

0631 

0733 

1444 

1545 

1647 

1748 

2457  2559 

2660 

2761 

633468  3569 

3670 

3771 

4^177 

4578 

4679 

4779 

5484 

5584 

5685 

5786 

6488 

6588 

6688 

6789 

7490 

7590 

7690 

7790 

8489 

8589 

8689 

8789 

9486 

9586 

9686 

9785 

640481 

0581 

0680 

0779 

1474 

1573 

/672 

1771 

2465 

2563 
3551 

2662 
3650 

2761 
3749 

643453 

4439 

4537 

4636 

4734 

5422 

5521 

5619 

6717 

6404 

6502 

6600 

6698 

7383 

7481 

7579 

7676 

8360 

8458 

8555 

8653 

9335 

9432 

9530 

9627 

650308 

0405 

0502 

0599 

1278 

1375 

1472 

1569 

2246 

2343 

2440 
3405 

2636 
3502 

653213 

3309 

4177 

4273 

4369 

4465 

5138 

5235 

5331 

5427 

6098 

6194 

6290 

6386 

7050 

7152 

7247 

7343 

8011 

8107 

8202 

8298 

8965 

9060 

9155 

9250 1 

9916 

..11 

.106 

.201 

680865 

0960 

10.55 

1150 

IS  13 

1907 

2002 

2096 

2494 

1  2603 

2711 

3577 

3686 

3794 

4058 

4766 

4874 

5736 

5844 

5051 

6811 

6919 

7026 

^8841  7991 

809S 

8954!  906 1 

9167 

..21 

.128 

.234 

1086 

1192 

1298 

2148 

2254 

2360 

3207 

3313 

3419 

4264 

4370 

4475 

6319 

5424 

5529 

6370 

6476 

6581 

7420 

7525 

7629 

8466 

8571 

8676 

9511 

9615 

9719 

0552J  0666 

0760 

1592 

1695 

1799 

2628 

2732 

2836 

3663 

3766 

3809 

4695 

4798 

4901 

67241  5827 

5929 

6751 

6853 

69561 

7776 

7878 

7980 

8797 

8900 

9002 

9817 

9919 

..2ll 

0835 

0936 

1038 

1849 

1961 

2052 

2862 

2963 

3064 

3872 

3973 

4074 

4880 

4981 

6081 

6886 

6986 

6087 

6889 

6989 

7089 

7890 

7990 

8090 

8888 

8988 

9088 

9885 

9984 

..84 

0879 

0978 

1077 

1871 

1970 

2069 

2860 

2959 
3946 

3058 

3847 

40441 

4832 

4931 

60291 

5816 

6913 

6011 

6796 

6894 

6992 

7774 

7872 

7969 

8750 

8848 

8345 

9724 

9821 

9919 

0696 

0793 

0890 

1666 

1762 

1859 

2633 

2730 

2826 

35981  3895 

3791 

4502 

4658 

4754 

5523 

6619 

5715 

6482 

6577  66731 

74.381  7534!  76291 

8393 

8488 

8584 

9346 

9441 

9536 

.296 

.391 

.486 

1245 

1339 

1434 

2101 

22S(> 

2330 

2819 
3902 
4982 
6059 
7133 
8205 
9274 
.341 
1405 
2466 


3525 
4581 
6634 
6686 


2928 
4010 
5089 
6166 
7241 
8312 
9381 
.447 
1611 
2572 


3630 
4686 
6740 
6790 


^734  7839 
8884 


8780 
9824 
0864 
1903 
2939 
39731 
50041 
60321 
70581 
80821 
9104 
.  123; 
1 1 39 1 
2153| 
3105 


9928 
0968 
2007 
3042 


4175 
5182 
6187 
7189 
8190 
9188 
.183 
1177 
2168 
3156 


4143 
5127 
6110 
7089 
8067 
9043 
..16 
0987 
1956 
2923 


3888 
4850 
.5810 
6769 
7725 
8879 
9631 
.581 
1629 
2475 


4076 
5107 
6136 
7161 
8185 
9206 
.224 
1241 
2256 
3206 
4276 
5283 
6287 
7290 
8290 
9287 
.283 
1276 
2267 
3255 
4242 
5226 
6208 
7187 
8165 
9140 
.113 
1084 
2053 
3019 


3036 
4118 
5197 
6274 
7348 
8419 
9488 
.554 
1617 
2678 
3736 
4792 
6845 
68951 
79431 
8989 
..32 
1072 
2110 
3JI46 
4179 
6210| 
6238 
72631 
8287 
93081 
.326 
1342 
2330 
3367 


3984 
4946 
.5906 
6864 
7820 
8774 
9726 
.6761 
1623 
2569 ! 


4376 
6383, 

6388 
7390 
8389i 
9387 
.382 
1375! 
2366 
3354 
4340 
5324 
6306 
7285 
8262 
9237 
.210 
1181 
2150 
3116 


4080 
5042 
60021 
69601 

79161 
8870( 
9821 
.7711 
17l8i 
2663! 


108 

u.s 

108 

107 

107 

107 

107 

100 

H)6 

1(!6 

106 

105 

106 

105 

106 

104 

101 

1(>4 

1(|4 

\03 

103 

103 

103 

102 

102 

102 

102 

101 

101 

100 

100 

100 

100 

99 

99 

99 

99 

99 

_99 

98 

98 

98 

98 

98 

97 

97 

97 

97 

97 

96 

96 

96 

96 

96 

95 

95 

96 

95 

96 


3  I  4 


6  I  7  I  8  I  9 


A  TAIILE  OF  I.'-»GARI'niMS  FKOM    1    TO   10.000. 


Jl-J 

U   i   I  i  2  i  3  1  4  1  5  1  6  1  7  1  8  i  9  !  I).  1 

4fi0 

6627581 

2852 1 2947 1  304 1 ,  3 1 35  3230,  3324,  34 1 8 1 35 1 2 1 3607 ,  94  1 

461 

3701 

37951 

38891 

3983  407814172 

4266  4360  44.54 

4.548! 

94 

462 

4642 

4736 

48301 

4924' 50181 5112 

.5206  5299  5393 

5487! 

94 

463 

5581 

5675 

5769! 

.58621  59.16  jMjSO 

614:3  62371  6331 

6424 

94 

464 

6518 

6612 

67051 

6799  6892 i  6986 

7079  717317266 

7360 

94 

165 

7453] 

7546 

7640J 

7733! 

7826:7920 

8013  810618199 

8293 

93 

466 

8386] 

8479 

8572  j 

8665 

8759188521 

8945  9038 1 9131 

9224 

93 

467 

9317 

9410 

9503! 

9596 

9689 i  9782! 

9875 

99671 ..60 

.1.53 

93 

468 

670246 

0339 

043 1  i 

0524 

06171 

0710 

0802 

0895  0988 

1080 

93 

469 

1173 

1265 

1358 

1451 

1.543, 

1636 

1728 

1821  1913 

2005 

9.3 

470 

672098 

2190 

2283 

23751 

2467 

2560 

2652 

2744  2836 

2929 

'92 

471 

1021 

3113 

3205 

3297 

3390 

3482 

C  74 

3666137.58 

38.50 

92 

472 

3942 

4034 

4126 

4218 

4310 

4402 

4494 

4.58614677 

4769 

•92 

473 

4861 

4953 

5045 

5137 

5228 

5320 

.5412 

.55031  .5.595 

5687 

92 

474 

5778 

5870 

5962 

6053 

6145 

6236 

6328 

6419 

6511 

6602 

92 

475 

6694 

6785 

6876 

6968 

7059 

7151 

7242 

7333 

7424 

7516 

91 

476 

7607 

7698 

7789 

7881 

7972 

8063 

8154 

8245 

8336 

8427 

91 

477 

8518 

8609 

8700 

8791 

8882 

8973 

9064 

91,55 

9246 

9337 

91 

478 

9428 

9519 

9610 

9700 

9791 

9882 

9973 

..63 

.154 

.245 

91 

479 

680336 

0426 

0517 

0607 

0698 

0789 

0879 

0970 

1060 

1151 

91 

480 

681241 

13{^2 

1422 

1.513 

1603 

1693 

"1 784 

1874 

1964 

2055 

90 

481 

2145 

2235 

2326 

2416 

2506 

2596 

2686 

2777 

2867 

2957 

90 

482 

3047 

3137 

3227 

3317 

3407 

3497 

3.587 

,3677 

3767 

3857 

90 

483 

3947 

4037 

4127 

4217 

4307 

4396 

4486 

4576 

4666 

4756 

90 

484 

4845 

4935 

5025 

5114 

5204 

.5294 

5383 

.5473  5563 

5652 

90 

485 

5742 

583 1 

5921 

6010 

6100 

6189 

6279 

63681  64.58 

6547 

89 

486 

6636 

6726 

6815 

6904 

6994 

7083 

7172 

7261 

7351 

7440 

89 

487 

7529 

7618 

7707 

7796 

7886 

7975 

8064 

81.53 

8242 

8331 

89 

488 

8420 

8509 

8598 

8687 

8776 

8865 

8953 

9042 

9131 

9220 

89 

489 

9309 

9398 

9486 

9575 

9664 

9753 

9841 

9930 

.,19 

.107 

89 

490 

690196 

0285 

0373 

0462 

0550 

0639 

0728 

0816 

0905 

0993 

89 

491 

1081 

1170 

1258 

i.347 

14.35 

1524 

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1700 

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88 

492 

1965 

2053 

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2759 

88 

493 

2847 

2935 

3023 

3111 

3199 

.3287 

33751 ,34631  3551 

3639 

88 

494 

3727 

3815 

3903 

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4078 

4166 

42.54 

4342 

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87 

497 

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3979 

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4151 

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4322 

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5008 

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1  53.50 

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5522 

5607 

5693 

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86 

508 

5864 

5949 

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513 

710117 

0202 

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0.540 

0625 

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85 

514 

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1132 

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1385 

1470 

1.554 

1639 

1723 

84 

615 

1807 

1892 

1976 

2060 

2144 

2229 

2313 

2397 

2481 

2566 

84 

516 

2650 

2734 

2818 

2902 

2986 

1  3070 

3154 

3238 

3323 

3407 

84 

517 

3491 

3575 

36.50 

3742 

!  3826 

13910 

3994 

4078 

4162 

4246 

84 

518 

4330 

4414 

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1  4665 

4749 

4833 

4916 

5000 

5084 

84 

519 

1   5167 

5251 

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!  .5502'  .'■^586 

5669'.57.53' .58361  5920 

84 

LiL 

1   0   1  1  1  2  1  3  1  4  1  5  1  6  1  7  1  8  !  9  1  dH 

A  TARLF.  OF  LOaAKITirSIS  FKOM   I   TO   10,000. 


N. 

1   0   1   1   1  2  1  3  1  4  1  5  1  6  i  7  1  «  1  W 

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716003.  6087.  G170 

6254 

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6504 

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6838 

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522 

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8419 

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8502 

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8668 

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8834 

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9000 

9083  9165 

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525 

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0490!  0573i  0655 

0738  0821 

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83 

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0986 

1068 

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1233 

1316 

1398  1481 

1663, 1646 

1728 

82 

527 

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2058 

2140 

2222  2305 

2.387  2469 

2552 

82 

528 

2634 

2716 

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2881 

2963 

3045  3127 
3866  3948 

3209  3291 

3374 

82 

529 

3456 

3538 

3620 

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4030!  4112 

4194 

82 

530 

724276 

4358 

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4522 

4604 

4685 

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5,585 

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81 

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539 

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79 

552 

1939 

2018 

2096 

2175 

2254 

2332  2411 

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2668 

2646 

79 

553 

2725 

2804 

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2961 

3039 

3118 

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3510 

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561 

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9040 

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9.504 

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77 

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77 

563 

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564 

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77 

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34.30 

3.506 

77 

567 

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4042 

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4196 

4272 

77 

558 

4348 

4425 

4501 

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46.54 

4730 

4807 

4883 

4960 

5036!  76  1 

569 
570 

5112 

5189 
5951 

5265 
6027 

5341 
6103 

5417 
6180 

5494 
6256 

5570 

5646 

5722 
6484 

5799 
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76 

76 

756875 

6332 

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6636 

6712 

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6864 

6940 

7016 

7092 

7168 

7244 

7320 

76 

572 

7396 

7472 

7548 

7624 

7700 

7775 

7851 

7927 

8003 

8079 

76 

573 

8155 

8230 

8306 

8382 

8458 

8533 

8609 

8685 

8761 

8836 

76 

574 

8912 

8988 

9063 

9139 

9214 

9290 

9366 

9^141 

9517 

9.592 

76 

575 

9668 

9743 

9319 

9894 

9970 

..45 

.121 

.196 

.272 

.347 

75 

576 

769422 

0498 

0573 

0649 

0724 

0799 

0875 

0950 

1025 

1101 

76 

577 

1176 

1251 

1326 

1402 

1477 

1562 

1627 

1702 

1778  18.53 

75 

578 

1928 

2003 

2078 

2153 

2228 

2303|  2378 

2463 

2529  2604 

75 

579 

2679 

27541  2829 

2904' v»i  8 

30631  3128' 3203 

3278  .3353'  75  1 

N. 

U   1  1   1  2  1  3   1  4  1  5  1  6  1  7  1  H  i  y   '  J).  1 

10 


A  TABLE  OP  LOGAKITirOTS  FROM   1  TO   10,000. 


N. 

1'  0   1   1  f  2  1  3  i  4  1  5  1  6  1  7  1  8  i  9  !  D. 

55o 

7()342fc 

i   350:j 

3a78r  3B53,  3727  3802,  3877,  39521  4027 1  4l0l|  75 

581 

417C 

4251 

4326!  4400| 4475  4550  46241  4691 

477414848!  .  75  i 

582 

492C 

;  4998 

5072  5147  5221  5296, 53701  544? 

55201  5594'  75  1 

583 

566ij 

5743 

5818,  5892  5966  6041 1  61 151  619' 

6264 

633S 

74 

584 

6412 

6487 

6562  66361 6710  67851 68591  6933  7007 

7082 

74 

585 

7156 

7230 

7304  7379  7453  7527|  7601  7675  774'J 

782? 

74 

586 

789S 

17972 

8046  8120  8194  826818342  841618490 

8564 

74 

587 

8633 

18712 

87861  8860  8934  9008|  9082i  9 J  56 
9525!  9599  9673i  97461  9820i  9894 
0263  0336  0410;'  0484!  0557|  0631 

9230 

9303 

74 

588 

9377 

I  9451 

9968 

..42 

74 

589 

770115 

0189 

0705 

107781  74  1 

5ii0 

770852 

!0926 

09991  1073  1146 

1220  1293!  1367 

1440 

1  i5"l4|  74  1 

591 

158? 

i  1661 

1734  1808 

1881 

1955i  2028'  2102 

2175 

2248 

73 

592 

2322 

!  2395 

24681 2542 

2615 

26881  27621  2835 

2908 

2981 

73 

593 

3055 

13128 

3201!  3274 

3348 

3421|  3494  3567 

3640 

3713 

73 

594 

3786 

3860 

3933 

4006 

4079! 4152!  42251  4298 

4371 

4444 

73 

595 

4517 

4590 

4663 

4736 

4809!  48821  4955!  5028 

5100 

5173 

73 

596 

5246 

5319 

5392 

5465 

!5538 

56101  5683.  575615829 

5902 

73 

597 

5974 

6047 

6120 

6193 

6265 

63331  641116483 

6556 

6629 

73 

598 

6701 

6774 

0848 

69i9 

6992 

7064! 7137 

7209 

7282 

7354 

73 

599 

7427 

i  7499 

7572 

7644 

7717 

7789|  7862 

7934 

8006 

8079 

[  72 

600 

77815118224 

8296 

8368 

8441 

85l3i  8585 

8658 

8730 

8802 

"72 

601 

8874  8947 

9019 

9091 

9163 

92361  9308 

9380 

9452 

9524 

72 

602 

9596  9669 

9741 

9813 

9885 

9957!  ..29 

.101 

.173 

.245 

72 

603 

780317  0389 

0461 

0533 

0605 

0677 

10749 

0821 

0893 

0965 

72 

604 

1037  1109 

1181 

1253 

1324 

1396 

1  1468 

1540 

1612 

16S4 

72 

605 

1755  1827 

1899 

1971 

2042 

2114 

i2186 

2258 

2329 

2401 

72 

606 

2473 i  2544 

2616 

26SS 

2759 

2831 

12902 

2974 

3046 

3117 

72 

607 

3189  3260 

3332 

34031 3475 

3546 

3618 

3689 

3761 

3832 

71 

608 

390413975 

4046,4118!  4189 

4261 

4332 

4403!  4475 

4546 

71 

609 

461714689 

4760 

4831!  4902 

4974 

5045 

5116|  5187 

5259 

71 

610 

7853301  5401 

5472 

6543 

5615 

5686 

5757 

582'8 

5899 

5970 

71 

611 

604116112 

6183 

6254 

6325^  6396 

6467 

6538 

6609 

6680!  71  1 

612 

67511 6822 

6893  6964 

7035! 7106 

7177 

7248 

7319 

7390 

71 

613 

74601 7531 

7602 

7673 

7744  7815 

78S5 

7956 

&027 

8098 

71 

614 

816818239 

8310 

8381 

8451!  8522 

8593 

8663 

8734 

8804 

71 

615 

88751 8946 

9016 

9087 

915719228 

9299 

9369 

9440 

9510 

71 

616 

958 1|  9651 

9722 

9792 

9863!  9933 

...4 

..74 

.144 

.215 

70 

617 

7902851 0356 

0426 

0496 

0567 

0637 

0707  0778 

0848 

0918 

70 

618 

0988  1059 

1129! 119& 

12691 

1340 

1410 

1480 

1550 

1620 

70 

619 

1691  1761 

183 1|  1901 

1971! 

2041 

2111 

2181 

2252; 

2322J  70  1 

620 

792392, 

2462 

2532!  2602|  2672! 

2742 

2812 

2882 

2952i 

3022 

70 

621 

3092 

3162 

3231 1 330 l! 

3371 

3441 

3511 

3581 

3651 

3721 

70 

622 

3790 

3860 

39301  4000| 

4070 

4139 

4209 

4279 

4349 

4418 

70 

623 

4488 

4558 
5254 
5949 

4627  4697 

4767  4836 

4906 

4976 

5045 

5115 

70 

624 

5185 

5324! 5393 

5463  5532 

5602 

5672 

5741 

5811 

70 

625 

5880 

60191  6088 

6158  6227 

6297 

6366 

6436 

6505 

62 

626 

6574! 

6644 

6713!  6782!  685216921 

6990  7060! 

7129 

7198 

69 

627 

7268 1 

7337 

7406!  7475!  ^545^  7614 

7683  7752 

7821 

7890 

69 

628 

7960; 

8029 

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8374  8443! 

8513 

8582 

69 

629 

865l| 

8720  i 

8789!  8858  8927J  8990 

9005i  91341 

9203! 

9272 

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630 

7993411 

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94781  95^^71  96 10  9685 

9754!  98231 

9892 

9961 

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631 

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0580 

06481  69 1 

632 

0717 

0786 

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1266 

13351 

69 

633 

1404 

1472| 

15411  1609i  J678  1747!  1815  18841  1952j  2021 

69 

634 

2039 

2158; 

2226! 2295| 2363  2432i 2500  2568! 2637  2705 

69 

635 

27741 

2842 

29l0i  2979!  3047  3116;  3184:  3252i  3321 1  3389 

68 

633 

3457! 

3525! 

3594!  3662!  3730  3798!  38071  3935i  4003'  4071 

68 

637 

41391 

42081 

42761  4344'  4412  4480'  4548,  46l6|  46S5|  47531 

68 

638 

48211 

4889^  4957!  5025  5093  5161J  52291  52'J7;  53651  543,5,'  68  1 

639  ' 

550  ll 

5.^09'  56371  5705  5773'  584l'  59031  59^6'  6044'  6!  12^  68 

N.  I 

0   I  1   1  2  1  3  1  4  !  5  1  6  1  7  1  8  1  9  i  D.  i 

A  TARLE  OF  LOGARITHMS  FROM    I  TO   10,000. 


11 


040' 

l_o_ 

1  ' 

1  2 

1  3_ 

i  4 

1  5 

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641 

6858 i  6926 

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7129 

7197 

7201!  7.332 

7400 

7467 

68 

612 

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76701  7738 

7806 

7873 

7941 

8008 

8076 

8143 

68 

643 

8211 

8279 

8346!  8414 

8481 

8549 

8616 

8684 

8751 

8818 

f)7 

644 

8886 

8953 

9021 

9098 

9156 

9223 

9290 

9358 

9425 

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67 

645 

9560 

9627 

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9762 

9829 

9896 

9964 

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67 

646 

8 1 0233 {0300 

0367 

0434 

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0904  0971 

1039 

1106 

1173 

12401 1307 

1374 

1441 

1.508 

67 

648 

1575  1642 

1709 

1776 

1843 

19I0I 1977 

2044 

2111 

2178 

67 

619 

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3181 

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3314 

3381  .3448 

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67 

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3848 

3914 

3981 

4048 

4114 

4181 

67 

652 

4248 

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4381 

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4,581 

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654 

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57 1 1 

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5843 

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5976 

6042 

6109 

6175 

66 

655 

6241 

6308 

6374 

6440 

6.506 

6573 

6639 

6705 

6771 

6838 

66 

656 

6904 

6970 

7036 

7102 

7169 

7235 

7301. 

7367 

7433 

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06 

657 

7565 

7631 

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7962 

8028 

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8160 

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658 

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8,358 

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8.556 

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8885 

895 1 

9017 

9083 

9149 

9215 

9281 

9.346 

9412 

9478 

66 

660 

819544 

9610 

9676 

9741 

9807 

9873 

9939 

...4 

..70 

.  136 

66 

661 

820201 

0267 

0333 

0399 

0464 

0530 

0.595 

0661 

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66 

662 

0858 

0924 

0989 

10.55 

1120 

1186 

1251 

1317 

1382 

1448 

66 

663 

1514 

1579 

1645 

1710 

1775 

1841 

1906 

1972 

2037 

2103 

65 

664 

2168 

2233 

2299 

2364 

2430 

2495 

2560 

2626 

2691 

27.56 

65 

665 

2822 

2887 

2952 

3018 

3083 

3148 

3213 

3279 

3344 

3409 

65 

666 

3474 

3539 

3605 

3670 

3735 

3800 

3865 

3930 

3996 

4061 

65 

667 

412(;' 

4191 

4256 

4321 

4386 

4451 

4516 

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65 

668 

4776 

4841 

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65 

069 

5426 

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5556 

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5880 

5945 

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65 

670 

82607^ 

6140 

6204 

6269 

6334 

6399 

6464 

6528 

6593 

6658 

65 

671 

6723 

6787 

6852 

6917 

6981 

7046 

7111 

7175 

7240 

7305 

05 

672 

7369 

7434 

7499 

7563 

7628 

7692 

7757 

7821 

7886 

7951 

65 

673 

8015 

8080 

8144 

8209 

8273 

8338 

8402 

8467 

8.531 

8595 

64 

674 

8660 

8724 

8789 

8853 

8918 

8982 

9046 

9111 

9175 

9239 

64 

675 

9304 

9368 

0432 

9497 

9561 

9625 

9690 

97.54 

9818 

9882 

64 

676 

9947 

..11 

..75 

.139 

.204 

.268 

.332 

•396 

.460 

.525 

64 

677 

830589 

0653 

0717 

0781 

0845 

0909 

0973 

1037 

1102 

1166 

64 

678 

1230 

1294 

13.58 

1422 

1486 

1,5,50 

1614 

1678 

1742 

1806 

64 

679 

1870 

1934 

1998 

2062 

2126 

2189 

2253 

2317 

2381 

2445 

64 

680 

832509 

2573 

2637 

2700 

2764 

2828 

2892 

29"56 

3020 

3083 

64 

681 

3147 

.3211 

3275 

3338 

3402 

3466 

3530 

3593 

3657 

3721 

64 

682 

3784 

3848 

3912 

3975 

4039 

4103 

4166 

4230 

4294 

4357 

64 

683 

4421 

4484 

4548 

4611 

4675 

4739 

4802 

48.66 

4929 

4993 

64 

684 

5056 

5120 

5183 

5247 

i.310 

.5373 

.5437 

5500 

5564 

5627 

63 

685 

5691 

5754 

.5817 

.5881 

5944 

6007 

6071 

6134 

6197 

6261 

63 

686 

6324 

6387 

6451 

6514 

6577 

6641 

6704 

6767 

6830 

6894 

63 

687 

6957 

7020 

7083 

7146 

7210 

7273 

7336 

7399 

7462 

7525 

63 

688 

7588 

7652 

7715 

7778 

7841 

7904 

7907 

8030 

8093 

81.56 

63 

689 

8219 

8282 

8345 

8408 

8471 

8534 

8597 

8660 

8723 

8786 

.63 

690 

838849 

8912 

8975 

9038 

9101 

9164 

9227 

92891  9352 

9415 

63 

691 

9478 

9541 

9604 

9667 

9729 

9792 

9855 

991819981 

..43 

63 

692 

840106 

0169 

0232 

0294 

0357 

0420 

0482 

05451  0608 

0671 

63 

693 

0733 

0796  0859 

0921 

0984 

1046 

1109 

1172! 1234 

1297 

63 

694 

1359 

1422  1485 

1.547 

1610 

1672 

1735 

17971  1860 

2422  2484 

1922 

63 

695 

1985 

2047  2110 

2172 

2235 

2297 

2360 

2.547 

62 

696 

2609 

2672  2734 

2796 

2859 

2921 

2983 

3046  3108 

3170  62 

697 

3233 

3295  3357 

3420 

3482 

3.544 

3606 

3669  3731 

3793  62 

69S 

3855 

3918  3980 

4042 

4104 

4166 

4229 

4291:43.53 

4415  62 

699 

4477 

4539  460: 

4«f)I 

4726 

4788 

48  5( 

4912  4974; 5036'  o2 ' 

1^1 

0   Il|2!3i4|5|6|7|sl9|n.  1 

16 


12 


A  TABLE  OF  L0GARITH3TS  FROM  1  TO  10,000. 


N.  1   0   1  1  1  2  1  3  1  4  1  5  1  fi  1  7  1  8  1  9  1  D.  1 

7t)() 

845098 

15160.6222 

5284 
5904 

5346.540815170 

5532 

5594 

5656 

02 

701 

5718 

5780 

5842 

5966 

6028  6090 

0151 

6213 

6275 

62 

702 

6337 

1  6399 

6461 

6523 

6585 

6646! 6708 

6770 

6832 

6894 

62 

703 

6955 

17017 

7079 

7141 

7202 

7264 

7326 

7388 

7449 

7511 

62 

704 

7573 

7634 

76  £6 

7758 

7819 

7881 

7943 

8004 

8066 

8123 

62 

705 

8189 

8251 

8312 

8374 

8435 

8497 

8559 

8620 

8682 

8743 

02 

706 

8805 

8866 

8928 

8989 

9051 

9112 

9174 

9235 

9297 

.y358 

61 

707 

9419 

9481 

9542 

9604 

9665 

9726 

9788 

9849 

9911 

9972 

61 

708 

850033 

0095 

0156 

0217 

0279 

0340 

0401 

0462 

0524 

0585 

61 

709 
710 

0046 

0707 
1320 

0769 
1381 

0830 
1442 

0891 
1503 

0952 
1564 

1014 
1625 

1075 
1686 

1136 
1747 

1197 

1809 

61 
61 

85125S 

711 

1870 

1931 

1992 

2053 

2114 

2175 

2236 

2297 

2358 

2419 

61 

712 

2480 

2541 

2602 

2663 

2724 

2785 

2846 

29^7 

2968 

3029 

61 

713 

3090 

3150 

3211 

3272 

3333 

3^9^ 

3455 

3516 

3577 

3637 

61 

714 

3898 

3759 

3820 

3881 

3941 

4002 

40' 3 

4124 

4185 

4245 

61 

715 

4306 

4367 

4428 

4488 

4549 

4610 

4670 

4731 

4792 

4852 

61 

716 

4913 

4974 

5034 

5095 

5156 

5216 

5277 

5337 

5398 

5459 

61 

717 

5519 

5580 

5640 

5701 

5761 

5822 

5882 

5943 

6003 

6064 

61 

718 

6124 

6185 

6245 

6306 

6366 

6427 

6487 

6548 

6608 

6668 

60 

719 

6729 

6789 

6850 

6910 

6970 

7031 

7091 

7152 

7212 

7272 

60 

720 

857332 

7393 

7453 

7513 

7574 

7634 

7694 

7755 

7815 

7875 

60 

721 

7935 

7995 

8056 

8116 

8176 

8236 

8297 

8357 

8417 

8477 

60 

722 

8537 

8597 

8657 

8718 

8778 

8833 

8398 

8958 

9018 

9078 

60 

723 

9138 

9198 

9258 

9318 

9379 

9439 

9499 

9559 

9619 

9679 

60 

724 

9739 

9799 

9859 

9918 

9978 

..33 

..93 

.1.58 

.218 

.278 

60 

725 

860333 

03&S 

0458 

0518 

0578 

0637 

0697 

0757 

0817 

0877 

60 

726 

0937 

0996 

1056 

1116 

1176 

1236 

1295 

1.355 

1415 

1475 

60 

727 

1534 

1594 

1654 

1714 

1773 

1833 

1893 

1952 

2012 

2072 

60 

728 

2131 

2191 

2251 

2310 

2370 

2430 

2489 

2549 

2608 

2688 

60 

729 
730 

2728 

2787 
3382 

2847 
3442 

2906 
3501 

2956 

3025 
3620 

3085 
3680 

3144 
"3739 

3204 
3799 

3263 

3858 

60 
59 

863323 

356 1 

731 

3917 

3977 

4036 

4096 

4155 

4214 

4274 

4333 

4392 

4452 

59 

732 

4511 

4570 

4830 

4689 

4748 

4808 

4867 

4926 

4935 

5045 

59 

733 

5104 

5163 

5222 

5282 

5341 

5400 

5459 

5519 

5578 

5637 

59 

734 

5696 

5755 

5814 

5874 

5933 

•5992 

6051 

6110 

6169 

6228 

59 

735 

6287 

6346 

6405 

6465 

6524 

6583 

6642 

6701 

6760 

6819 

59 

73b 

6878 

6937 

6996 

7055 

7114 

7173 

7232 

7291 

7350 

7409 

59 

737 

7467 

7526 

7585 

7644 

7703 

7762 

7821 

7880 

7939 

7998 

59 

738 

8056 

8115 

8174 

8233 

8292 

8350 

8409 

8468 

8527 

8586 

59 

739 

740 

8644 

8703 
9290 

8762 

8821 
9408 

8879 
9466 

8938 
95-^5" 

8GJ7 
9584 

9056 
9642 

9114 
9701 

9173 

976;) 

59 
59 

869232 

9349 

741 

9818 

9377 

9935 

9994 

..53 

.111 

.170 

.228 

.287 

.345 

59 

742 

870404 

0462 

0521 

0579 

063S 

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U755 

0813 

0872 

0930 

58 

743 

0989 

1047 

1106 

1164 

1223 

1281 

1339 

1398 

1456 

1515 

58 

744 

1573 

1631 

i690 

1748 

1806 

1865 

1923 

1981 

2040 

2098 

58 

745 

2156 

2215 

2273 

2331 

2389 

2448 

2506 

2564 

2622 

2681 

58 

746' 

2739 

2797 

2855 

2913 

2972 

3030 

3088 

3146 

3204 

3262 

58 

747 

3321 

3379 

3437 

3495 

3553 

3611 

3669 

3727 

3785 

3344 

58 

748 

3902 

3960 

4018 

4076 

4134 

4192 

4250 

4308 

4366 

4424 

58 

749 

4482 

4540 

4598 

4656 

4714 

4772 

4830 

4888 

4945 

5003 

58 

750 

875061 

5119 

5177 

5235 

5293 

5351 

5409 

5466 

5524 

.5582 

58 

751 

5640 

5698 

5756 

5313 

5871 

5929 

5987 

6045 

6102 

6160 

58 

•752 

6218 

6276 

6333 

6391 

6449 

6507 

6564 

6622 

6680 

6737 

58 

753 

6795 

6853 

6910 

6968 

7026 

7083 

7141 

7199 

7256 

7314 

58 

754 

7371 

7429 

7487 

7544 

7602 

7659 

7717 

7774 

78;?2 

7889 

58 

755 

7947 

8004 

8062 

8119 

8177 

82:4 

8292 

8349 

8407 

8464 

57 

756 

8522 

8579 

8637 

8694 

3752 

8309 

8866 

8924 

8931 

9039 

57 

757 

9096 

9153 

9211 

9268 

9325 

9333 

9440 

9497 

9555 

9612 

57 

T^Q 

9669 

9726 

9784 

9841 

9398 

9956 

..13 

.70 

.127 

.185 

57 

759 

880242 

0299 

0356 

0113 

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0585 

.642 

0699 

0756 

57 

Ji. 

0   1  I  i  2.  1  3  1  4  1  5  1  6  1  7  i  a  1  9  1  D.  1 

A  TABI.F  OF  LOGAItlTIIMS  FR03I   1  TO   10,000. 


13 


N. 

0   1  1  !  2  1  3  1  4  1  5  1  6  i  7  1  8  i  9  1  D. 

Wo' 

880814,0871, 

0928  09851 

1042 

1099 

1 166 

I213j  12711 

1328.  57 

701 

1385 

1442 

1499 

1556 

1613 

1670 

1727 

1784! 

1841 

1898!  57 

703 

1955 

2012 

2069 

2126 

2183 

2240 

2297 

2354 

2411 

24681  57 

763 

2525 

2581 

2638 

2695 

2752 

2809 

2866 

2923 

2980130371  57 

7(54 

3093 

3160 

3207 

3264 

3321 

3377 

3434 

3491 

3548! 3605  57 

765 

3661 

3718 

3775 

3832 

3888 

3945 

4002 

4059 

4115U172'  57 

766 

4229 

4285 

4342 

4399 

4455 

4512 

4569 

4625 

46821  4739 

57 

767 

4795 

4852 

4909 

4965 

5022 

5078 

6135 

5192 

52481 5305 

57 

768 

5361 

5418 

5474 

5531 

5587 

5644 

5700 

5757 

5813, 6870 

57 

769 

770 

5926 
886491 

5983 
6547 

6039 

6096  ■ 
6660 

6152 
6716 

6209 
6773 

6265 
6829 

6321 

6885 

63781  6'i34 
6942! 6998 

56 
56 

6604 

771 

7054 

71111 

7167 

7223 

7280 

7336 

7392 

7449 

7505 1  7561 

56 

772 

7617 

7674 

7730 

7786 

7842 

7898 

7955 

8011 

8067  8123 

56 

773 

8179 

8236 

8292 

8348 

8404 

8460 

85  J  6 

8573 

8629  8685 

56 

774 

8741 

8797 

8853 

8909 

8965 

9021 

9077 

9134 

9190  9246 

56 

775 

9302 

9358 

9414 

9470 

9526 

9582 

9638 

9694 

97501 9806 

56 

776 

9862 

9918 

9974  ..30| 

..86 

.141 

.197 

.253 

.309  .365 

56 

777 

890421 

0477 

0533 

0589 

0645 

0700 

0756 

0812 

0868  0924 

56 

778 

0980 

1035 

1091 

1147 

1203 

1259 

1314 

1370 

1426  1482 

66 

779 

1537 

1593 

1049 

1705 

1760 

1816 

1872 

1928 

1983  2039 

56 

780 

892095 

2150 

2206 

2262 

2317 

2373 

2429 

2484 

2540  2595 

~56 

781 

2651 

2707 

2762 

2818 

2873 

2929 

2985 

3040 

3096  3151 

56 

782 

3207 

3262 

3318 

3373 

3429 

3484 

3540 

3595 

3651 

3706 

56 

783 

3762 

3817 

3873 

3928 

3984 

4039 

4094 

4150 

4205 

4261 

55 

784 

4316 

4371 

4427 

4482 

4538 

4593 

4648 

4704 

4759 

4814 

55 

785 

4870 

4925 

4980 

5036 

5091 

5146 

5201 

5257 

5312 

5367 

65 

786 

5423 

5478 

5533 

5588 

5644 

5699 

5754 

5809 

6864 

6920 

55 

787 

5975 

6030 

6085 

6140 

6195 

6251 

6306 

6361 

6416 

6471 

55 

788 

6526 

6581 

6636 

6692 

6747 

6S02 

6857 

6912 

6967 

7022 

55 

789 
790 

7077 

7132 
7682 

7187 

7242 
7792 

7297 

7847 

7352 
7902 

7407 
7957 

7462 
8012 

7517 
8067 

7572 
8122 

55 

55 

897627 

7737 

791 

8176 

8231 

8286 

8341 

8396 

8451 

8506 

8561 

8616 

8670 

55 

792 

8725 

8780 

8835 

8890 

8944 

8999 

9054 

9109 

9164 

9218 

65 

793 

9273 

9328 

9383 

9437 

9492 

9547 

9602 

9656 

9711 

9766 

55 

794 

9821 

9875 

9930 

9985 

..39 

..94 

.149 

.203 

.258 

.312 

65 

795 

900367 

0422 

047G 

0531 

0586 

0640 

0695 

0749 

0804 

0869 

55 

79G 

0913 

0968 

1022 

1077 

1131 

1186 

1240 

1295 

1349 

1404 

55 

797 

1458 

1513 

1567 

1622 

1676 

1731 

1785 

1840 

1894 

1948 

54 

798 

2003 

2057 

2112 

2166 

2221 

2275 

2329 

2384 

2438 

2492 

54 

799 
800 

2547 

2601 
3144 

2655 
3199 

2710 

2764 
3307 

2818 
3361 

2873 
3416 

2927 
3470 

2981 
3524 

3036 
3578 

64 
64 

903090 

3253 

801 

3633 

3687 

3741 

3795 

3849 

3904 

3958 

4012 

4066 

4120 

54 

802 

4174 

4229 

4283 

4337 

4391 

4445 

4499 

4553 

4607 

4661 

64 

803 

4716 

4770 

4824 

4878 

4932 

4986 

5040 

5094 

5148 

6202 

54 

804 

5256 

5310 

5364 

5418 

5472 

5526 

5580 

56.34 

5688 

6742 

54 

805 

5796 

5850 

5904 

5958 

6012 

6066 

6119 

6173 

6227 

6281 

54 

806 

G335 

6389 

6443 

6497 

6551 

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6658 

6712 

6766 

6820 

54 

807 

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6927 

6981 

7035 

7089 

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7304 

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64 

80ri 

7411 

7465 

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7626 

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7734 

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54 

809 
810 

7949 
908485 

8002 
8539 

8056 

8110 

8163 

8217 
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8270 
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8914 

8431 
8967 

•  64 
54 

8592  8646 

8699 

8860 

811 

9021 

9074 

9128  9181 

9235 

9289 

9342 

9396 

9449 

9503 

54 

812 

9556 

9610 

966319716 

9770 

9823 

9877 

9930 

9984 

..37 

53 

813 

910091 

0144 

0197 

0251 

0304 

0358 

0411 

0464 

0518 

0571 

53 

814 

0624 

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0731 

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0838 

0891 

0944 

0998 

1051 

1104 

63 

815 

1158 

1211 

1264 

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1371 

1424 

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1530 

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1637 

53 

816 

1690 

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1797 

1850 

1903 

1956 

2009 

2063 

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53 

817 

2222 

12275 

2328 

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3231 

53 

82^9 

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i  3337 

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3443 

3496 

3549 

1  3002 

3655! 3708 

376  ll  53  1 

^nT 

1   0   1  1  1  2  1  3  1  4  !  5  1  6  1  7  i  8  i  9  1  O. 

1 

14 


A  TAnLE  OF  IOnAl?ITir3IS  FK031    I  TO   10,000. 


N. 

(   0   1   1  !  2  I  3  1  4  1  5  1  G  1  7  1  8  i  9  1  D.  1 

S2() 

913314,  38871  3920|  3973,  4026 

4079 

(4132  41S4|4237 

I  4290r'63 

821 

4343  4396 

1 4449 j  4502 

4555 

4608 

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1  4713 

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822 

4872  4925 

1  4977!  5030 

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1  5241 

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53 

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55051 5558 

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5927  5980:  G033!  6035 

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7085  7138 

7190 

7243 

7295 

7348 

7400 

7453 

1  53 

827 

7506 

1  7558 

761 1|  7663 

7716 

7768 

7820 

78731 7925 

7978 

52 

828 

8030 

8083 

813518188 

8240 

8293 

8345 

1  8397i  8450 

8502 

62 

829 
830 

8555 
919078 

8607 
9130 

8659|8712 
91831  9235 

8764 
9287 

8816 

8869 

8921 
9444 

8973 
9496 

9026 
9549 

62 
'.02 

9340 

9392 

831 

9601 

9653 

9706 j  9758 

9810 

9862 

9911 

9967 

..19 

..71 

52 

832 

920123 

0176 

0228  0280 

0332 

0384 

0438 

0489 

0541 

0593 

52 

833 

0645 

0697 

0749  0801 

0853 

0906 

0958 

1010 

1062 

1114 

52 

834 

1166 

1218 

1270  1322 

1374 

1426 

1478 

J. 530 

1582 

1634 

52 

835 

1686 

1738 

1790|  1842 

1894 

1946 

1998 

2050 

2102 

2154 

52 

836 

2206 

2258 

231012362 

2414 

2466 

2518 

2570 

2622 

2671 

52 

837 

2725 

2777 

2829  2881 

2933 

2985 

3037 

3089 

3140 

3192 

62 

838 

3244 

3296 

3348  3399 

3451 

3503 

3555 

3607 

3658 

3710 

52 

839 
840 

3762 

3814 
4331 

3865  3917 
4383! 4434 

3969 
4486 

4021 
4533 

4072 

4589 

4124 
4641 

4176 
4693 

4228 
4744 

52 
52 

924279 

84  J 

4796 

4848 

4899 

4951 

.5003 

5054 

5108 

5157 

5209 

5261 

62 

842 

5312 

5364 

5415 

5467 

5518 

5570 

5621 

5673 

5725 

6776 

.52 

843 

5828 

5879 

5931 

5982 

6034 

6085 

6137 

6188 

6240 

6291 

61 

844 

6342 

6394 

6445 

6497 

6648 

6800 

6661 

67021  67.54 

6805 

51 

845 

6857 

6908 

6959} 7011 

7062 

7114 

7165 

7216J7268 

7319 

61 

846 

7370 

7422 

7473  7524 

7576 

7627 

7678 

7730 

7781 

7832 

6L 

847 

7883 

7935 

7986  8037 

8088 

8140 

8191 

8242 

8293 

8.345 

51 

848 

8396 

8447 

8498  8549 

8601 

8652 

8703 

8754 

8805 

8857 

51 

849 
850 

8908 
929419 

8959 
9470 

9010|9061 

9112 
9623 

9163 

9215 
9725 

9266 

9317 
9827 

9368 
9879 

51 
51 

9521 

9572 

9674 

9776 

851 

9930 

9981 

..32 

..83 

.1.34 

.185 

.236 

.287 

.333 

.389 

51 

852 

930440 

0491 

0542 

0592 

0643 

0694 

0745 

0796 

0847 

0898 

51 

853 

0949 

1000 

1051 

1102 

1153 

1204 

1254 

1305 

1356 

1407 

51 

854 

1458 

1509 

1560 

1610 

1661 

1712 

1763 

1814 

1865 

1915 

51 

855 

1966 

2017 

2068 

2118 

2169 

2220 

2271 

2322 

2372 

2423 

51 

856 

2474 

2524 

2575 

2626 

2677 

2727 

2778 

2829 

2879 

2930 

51 

857 

2981 

3031 

3082 

3133 

3183 

3234 

3285 

3335 

.3386 

3437 

51 

858 

3487 

3538 

3589 

3639 

3690 

3740 

.3791 

3841 

3892 

3943 

51 

859 
860 

3993 

4044 
4549 

4094 
4599 

4145 
4650 

4195 
4700 

4246 
4751 

4296 
4801 

4347 
4852 

4397 

4448 
49.53 

51 
60 

934498 

4902 

861 

5003 

5064 

5104 

5154 

5205 

5255 

5306 

5356 

.5406 

.5457 

50 

862 

5507 

5558 

6608 

5658 

5709 

6759'  6809 

5860 

.5910 

5960 

60 

863 

6011 

6061 

6111 

6162 

6212 

6262' 6313 

6363 

6413 

C463 

60 

864 

6514 

6564 

.6614 

6665 

6715 

6765 

6815 

6865 

6916 

6966 

50 

865 

7016 

7066 

7117 

7167 

7217 

7267 

7317 

7367 

7418 

7468 

60 

8661 

7518 

7568 

7618 

7668 

7718 

77J69 

7819 

7869  7919 

7969 

60 

867 

8019 

8069 

8119 

8169 

8219 

8269 

8320 

8370  8420 

8470 

50 

868 

8520 

8570 

8620 

8670 

8720 

8770 

8820 

8870  8920 

8970 

50 

869 

9020 

9070 

9120 

9170 

9220 

9270 

9320 

9369  9419 

9469 

50 

870 

9395-9 

9569 

9619 

9669 

9719 

9769 

9819 

9869 

9918 

9968 

60 

871 

940018 

0068 

0118 

0168 

0218 

0267 

0317 

0367 

0417 

0467 

50 

872 

0516 

0566 

0616 

0666 

0716 

0765 

0815 

0865 

0915 

0964 

50 

873 

UH4 

1064 

1114 

1163 

1213 

1263 

1313 

1362 

1412 

1462 

50 

874 

1511 

1561 

1611 

1660 

1710 

1760 

1809 

1859 

iq09 

1958 

50 

875 

2008 

2058 

2107 

2157 

2207 

2256 

2306 

2355 

2405 

2455 

50 

876 

2504 

2554 

2603 

2653 

2702 

2752 

2801 

28511 

2901 

2950 

50 

877 

3000 

3049 

3099 

3148 

3198 

3247 

3297 

33461  33961  3445  i 

49 

878 

3495 

35-44 

3593 

3643 

3692 

3742! 

3791 

3841  3890i 3939] 

49 

879 

3989 

4038  4088' 4137 

41861  42361  4285 

43351 4334' 4433  49  | 

N. 

0   i  1   1  2  1  8  1  4  1  5  '  6  1  7  1  8  '  9  1  D.  1 

A  TABLE  Cil'-  LOOAKTTIIMS  FROM  T  TO  10,000. 

1^ 

N. 

0   1  1  1  2  1  3  1  4  i  5  1  6  1  7  1  8  1  9  ;  l>.  1 

88b" 

944483  4532 

45 SI  46311 

4680 

4729 

4779 

48ii8i4877i49--;71  49  1 

881 

4976 

5025 

5074 

5124 

5173 

5222 

5272 

5321  .5370 

54191  49 

882 

5469 

5518 

5567 

5616 

5665 

5715 

5764 

.581315862 

59 1 21  49 

883 

5981 

6010 

6059 

6108 

6157 

6207 

6256 

6305] 0354 

6403!  49 

884 

6452 

C501 

6551 

6600 

6649 

6698 

6747 

6796 1 6845 

6894  49 

88.'i 

6943 

6992 

7041 

7090 

7140 

7189 

7238 

72871 7336 

7385 

49 

886 

7434 

7483 

7532 

7581 

7630 

7679 

7728 

7777  7826 

7875 

49 

887 

7924 

7973 

8022 

8070 

8119 

8168 

8217 

8266 

8315 

8364 

49 

888 

8413 

8462 

8511 

8560 

8609 

8657 

8706 

8755 

8804 

8853 

49 

889 

8902  8951 

8999 

9048 

9097 

9146 

9195 

9244 

9292 

9341 

49 

890 

949390 

94nyi 4488 

9536 

9585 

9634 

9683 

9731 

9780 

9829  49 1 

891 

9878 

99/6:9975 

..24 

..73 

.121 

.170 

.219  .2671 

.316 

49 

892 

950365 

04  4; 0462 

0511 

0560 

0608 

0057 

0706 

0754 

0803 

49 

893 

08*=: 

0.>00 

0949 

0997 

1046 

1005 

1143 

1192 

1240 

1289 

49 

894 

13J8  1386 

1435 

1483 

1532 

1.580 

1629 

1677 

1726 

1775 

49 

8^5 

1823 

18:2 

1920 

1969 

2017 

2066 

2114 

2163 

2211 

2260 

48 

d96 

2308 

2356 

2405 

2453 

2502 

2550 

2599 

2647 

2696 

2744 

48 

897 

2792 

2841 

2889 

2938 

2986 

3034 

3083 

3131 

3180 

3228 

48 

898 

3276 

3325 

3373 

3421 

3470 

3518 

3566 

3615 

3663 

3711 

48 

899 

3760 

3808 

3856 

3905 

3953 

4001 

4049 

4098 

4146 

4194 

48 

9U0 

954243 

4291 

4339 

4387 

4435 

4484 

4532 

4580 

4628 

4677 

ll8 

901 

4725 

4773 

4821 

4869 

4918 

4966 

.5014 

5062  5110 

5158 

48 

902 

5207 

5255 

5303 

5351 

5399 

5447 

5495 

5543  5592 

5640 

48 

903 

5688 

5736 

5784 

5832 

5880 

5928 

5976 

6024  6072 

6120 

48 

904 

6168 

6216 

6265 

6313 

6361 

6409 

6457 

6505  6553 

6601 

48 

905 

6649 

6697 

6745 

6793 

6840 

6888 

6936 

6984 

7032 

7080 

48 

go*? 

7128  7176 

7224 

7272 

7320 

7368 

7416 

7464 

7512 

7559 

48 

907 

7607  7655 

7703 

7751 

7799 

7847 

7894 

7942 

7990 

8038 

48 

908 

8086 

8134 

8181 

8229 

8277 

8325 

8373 

8421 

8468 

8516 

48 

909 

8564 

8612 

8659 

8707 

8755 

8803 

8850 

8898 

8946 

8994 

48 

910 

959041 

9089 

9137 

9185 

9232 

9280 

9328 

9375 

9423 

9471 

48 

911 

9518 

9566 

9614 

9661 

9709 

9757 

9804 

9852 

9900 

9947 

48 

912 

9995 

..42 

..90 

.138 

.185 

.2.33 

.280 

.328 

.376 

.423 

48 

"913 

960471 

0518 

0566 

0613 

0661 

0709 

0756 

0804 

0851 

0899 

48 

914 

0946 

0994 

1041 

1089 

1136 

1184 

1231 

1279 

1326 

1374 

47 

915 

1421 

1469 

1516 

1563 

1611 

1658 

1706 

1753 

1801 

1848 

47 

916 

1895 

1943 

1990 

2038 

2085 

2132 

2180 

2227 

2275 

2322 

47 

917 

2369 

2417 

2464 

2511 

2559 

2606 

2053 

2701 

2748 

2795 

47 

918 

2843 

2890 

2937 

2985 

3032 

3079 

3126 

3174 

3221 

3268 

47 

919 

3316 

3363 

3410 

3457 

3504 

3552 

3.-99 

3646 

3693 

3741 

47 

920 

963788 

3835 

3882 

3929 

3977 

4024 

4071 

4118 

4165 

^212 

47 

921 

4260 

4307 

4354 

4401 

4448 

4495 

4542 

4590 

4637 

46841  47  1 

922 

4731 

4778 

4825 

4872 

4919 

4966 

.''-013 

5061 

5108 

51.55 

47 

923 

5202 

5249 

5296 

5343 

5390 

5437 

5484 

5531 

55781 5625 

47 

924 

5672 

5719 

5766 

5813 

5860 

5907 

6954 

6001 

6048 

6095 

47 

925 

6142 

6189 

6236 

6283 

6329 

6376 

6423 

6470 

6517 

6564 

47 

926 

6611 

6658 

6705 

6752 

6799 

6845 

6892 

6939 

6986 

7033  47 

927 

7080 

7127 

7173 

7220 

7267 

7314 

7361 

7408 

7454 

7501  47 

928 

7548 

7595 

7642 

7688 
8156 

7735 

7782 

7829 

7875 

7922 

7969  47 

929 
930 

8016 

8062 
8530 

8109 
8576 

8203 

8249 

8296 
8763 

8343 

8810 

8390 
8856 

8430 
8903 

47 
47 

96S483 

8623!  8670i  8716 

931 

8950 

8996 

9043 

9090 

91.36  9183 

9229 

9276 

9323 1 9369 

47 

932 

9416 

9463 

9509 

9556 

9602  9649 

9695 

9742 

978919835 

47 

933 

9882 

9928 

9975 

..21 

..68  .114 

.161 

.207 

.254 

.300 

47 

934 

970347 

0393 

0440 

0486 

05.33  0579 

0626 

0672 

0719 

0765 

46 

935 

0S12 

0858 

0904 

0951 

0997  1044 

109011137 

1183 

1229 

46 

936 

1276 

1322 

1,369 

1415 

1461  1.508!  15.541  1601 

1647 

1693 

40 

937 

1740 

1786 

1832 

1879 

1925  19711  2018 

2064 

2110 

2157!  461 

933 

2203 

2249 

2295 

2342 

238812434!  2481 

2527 

257312619!  46  | 

_939_ 

266612712 

'  2758'  2804'  2851'  2897'  2943'  2989'  3035'  3082  46  | 

N. 

1   0   1  1  1  2  1  3 

I   ^•-. 

1  5  !  .6  i  7  1  8  I  9  1  0.  1 

'16  ~ 

16 

A  TABLE  OF  LlWJAKITIi:«S  FROM  1 

TO  10,000 

0   1  I  1  2  1  3  I  4  1  5  1  6  1  7  1  8  1  9  1  D.  1 

9i0| 

97312813174132201 

3266  331313359!  34051 

3451  34971 

35431 

46 

941 

3590 

3636! 36S2 

3728 

3774  3820] 

3866 

3913 

3959 

4005 

46 

942 

4051 

409714143 

4189 

4235 

4281 

4327 

43  ?4 

4420 

4466 

46 

943 

4512 

45581 4604 

4650 

4696 

4742 

4788 

4S34 

4880 

4926 

46 

944 

4972 

5018  5064 

5110 

5106 

5202 

5248 

.5294 

5340 

5386 

46 

945 

5432 

5478  5524 

5570 

5616 

5662 

5707 

5753 

5799  58451 

46 

916 

5891 

59371  5983 

6029  60751 

6121 

6167 

6212 

6258 

6ii04 

46 

917 

6350  6396!  6442| 

6488 

6533 

6579 

6625  66711 

6717 

6763 

46 

918 

6808 

6854  6900 

6946 

6992 

7037 

7083!  712917175 

7220 

46 

949 

7266 

7312  7358 

7403 

7449 

7495 

7541 

75861 7632 

7678 

46 

9ol) 

977724 

7769  7815 

786  r 

790G 

71.52 

7993 

8043  8089 

81.35 

'46 

951 

8181 

8226  8272 

8317 

8363 

8409 

8454 

8500  8540 

8591 
9047 

46 

952 

8637 

8683  8728 

8774 

8819 

8865 

8911 

8956  9002 

4a 

953 

9093 

9138 

9184 

9230 

9275 

9321 

9366 

9412  9457 

9503  46 1 

954 

9548 

9594 

9639 

9685  9730 

9776 

9821 

9867  9912 

9958 

46 

955 

980003 j  00491 

0094 

0140 

0185 

0231 

0276 

0322 

0367 

0412 

45 

956 

0458 

0503 

0549 

0594 

0640 

0685 

0730 

0776 

0821 

0867 

45 

957 

0912 

0957 

1 003 

1048 

1093 

1139 

1184 

1229 

1275 

1320 

45 

958 

1366 

1411 

1456 

1501 

1547 

1592 

1637 

1683 

1728 

1773 

45 

959 
960  1 

1819 

1864 
2316 

1909 
2362 

1954 
2407 

2000 

2045 
2497 

2090 
2543 

2135  2181 
258^  2633 

2226 
2678 

45 
45 

982271 

2452 

961 

2723 

2769 

2814 

2859 

2904 

2949 

2994 

304C 

3085 

3130 

45 

962 

3175 

3220 

3265 

3310 

3356 

3401 

3446 

3491 

3536 

3581 

45 

963 

3626 

3671 

3716 

3762 

3807  3852i  3897 

3942 

3987 

4032 

45 

961 

4077 

4122 

4167 

4212 

4257;  4302 

4347 

4392 

4437 

4-182 

45 

965 

4527 

4572 

4617 

4062 

4707 

4752 

4797 

4842 

4887 

4932 

45 

966 

4977 

5022 

5067 

5112 

51.57 

5202 

5247 

5292 

5337 

5382 

45 

967 

5426 

5471 

5516 

5561 

5606 

5651 

5696 

5741 

5786 

.5830 

45 

968 

5875 

5920 

5965 

6010 

6055 

6100 

6144 

6189 

6234 

0279 

45 

969 
970 

6324 

6369 

6413 
6861 

6458 
6906 

6503 
6951 

6548 
6996 

6593 
7040 

6637 

7085 

6682 
7130 

6727 
7175 

45 
45 

986772 

6817 

971 

7219 

7264 

7309 

7353 

7398 

7443 

7488 

7532 

7577 

7622 

45 

972 

7666 

7711 

7756 

7800 

7845 

7890 

7934 

7979 

8024 

8068 

45 

Q73 

81  13 

8157 

8202 

8247 

8291 

8336 

8381 

8425 

8470 

8514 

45 

074 

8559 

8604 

8648 

8693 

8737 

8782 

8826 

8871 

8916 

8960 

45 

975 

9005 

90491  9094 

9138 

9183 

9227 

9272 

,9316 

9.361 

9405 

45 

976 

9450 

9494 

9539 

9583 

9628 

9672 

9717 

9701 

9806 

9850 

44 

977 

9895 

9939 

9983 

..28 

..72 

.117 

.161 

.206 

.2.50 

.294 

44 

978 

990339 

0383 

0428 

0472 

0516 

0.561 

0005 

0650 

0694 

0738 

44 

979 
98^ 

0783 
991226 

0827 

0871 

0916 
1359 

0960 
1403 

1004 
1448 

1049 
1492 

1093 
1536 

1137 
1580 

1182 
1625 

44 
44 

1270 

1315 

981 

1669 

1713 

1758 

1802 

1846 

1890 

1935 

1979 

2023 

2067 

44 

982 

2111 

2156 

2200 

2244 

2288 

2333 

2377 

2421 

2465 

2509 

44 

983 

2554 

2598 

2642 

2686 

2730 

2774 

2819 

2863 

2907 

2951 

44 

984 

2995 

3039 

3083 

3127 

3172 

3216 

3260 

3304 

3348 

33921  44 1 

985 

3436 

3480 

3524 

3568 

3613 

3657 

3701 

3745 

3789 

3833 

44 

986 

'   3877 

3921 

3965 

4009 

4053 

4097 

4141 

4185 

4229 

4273 

44 

987 

4317 

4361 

4405 

4449 

4493 

4537 

4581 

4625 

4669 

4713 

44 

988 

4757 

4801 

4845 

4889 

4933 

4977 

5021 

.5065 

5108 

51.52 

44 

989 
990 

5196 

5240 
5679 

6284 

5328 

5372 
.5811 

5416 
5854 

5460 
.5898 

5504 
5942 

5547 
5986 

5.591 
6030 

44 
44 

995635 

5723 

5767 

991 

6074 

6ir< 

6161 

6205 

6249 

6293 

6337 

6380 

6424 

64681  44 1 

992 

6512 

6555 

6599 

6643 

6687 

6731 

6774 

6818 

6862 

6906 

44 

l;93 

6949 

6993 

7037 

7080 

7124 

7168 

7212 

7255 

7299 

7343 

4-1 

094 

7386!  7430 

7474 

7517 

7.561 

7605!  7648 

7692 

77.3C 

7779 

44 

P')5 

7823 

7867 

7910 

7954 

7998 

8041 

8085 

8129 

8172 

8216 

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8259 

8303 

8347 

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8434 

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8521 

8564 

8603 

8652 

44 

997 

8695 

8739 

8782 

882018869 

8913 

8956 

9000 

9043 

9087 

44 

998 

9131 

9174 

9218 

90j1  9305 

9348 

9392 

9435 

9479 

9522 

44 

939 

9565 

9'>09 

9R5'2l  9Hnn|973!tl  9783 

9826 

9870 

99131  99  >7 

43 

N. 

1   0   1  1  1  2  1  3  1  4  1  5  1  6  1  7  1  8  1  o  '  D.  1 

A  TABLE 


OF 

LOGARITHMIC 
8INES    AND    TANGENTS 

FOR  EVERT 

DEGREE  AND  MINUTE 

OF  THE  QUADRANT. 


N.  B  The  minutes  in  the  left-hand  column  of  each  page, 
increasing  downwards,  helong  to  the  degrees  at  the  top ;  and 
those  increasing  upwards,  in  the  right-hand  column,  belong  to 
the  degrees  below. 


18 

(0  Tegree.)  a  table  of  logarithmic 

M. 

1   Sine 

1   1). 

1  cnsi,...  |n. 

I   Ta,.!:.   1   1). 

1   <,"ni:uu.  1   1 

"T 

O.OihJOJO 

10.000000 
000000 

0.000000 

lllii:llll:. 

"60 

1 

G. 463726 

.501  717 

00 

6.463726 

501717 

13.. 536274 

59 

2 

764756 

293485 

000000 

00 

764756 

293483 

235244 

58 

3 

940817 

208231 

000000 

00 

940847 

208231 

0591.53 

57 

4 

7.065786 

161517 

000000 

00 

7.06578H 

161517 

12.934214 

56 

f) 

162696 

13198S 

OJOJOO 

00 

162696 

131939 

83/304 

55 

6 

241877 

111.575 

9.99');)99 

01 

241878 

1 1 1 578 

7.58122  54 

7 

308824 

966.53 

999999 

01 

308825 

996.53 

691175  53 

8 

366816 

85254 

999999 

01 

.366817 

852.54 

633183  52 

9 

417968 

76263 

999999 

01 

417970 

76263 

6820301  51 

10 
il 

463725 
7.5U5118 

68988 
'62981 

999998 

01 
01 

463727 

68988 

.536273 

50 

9.999998" 

7.505120 

62981 

12.494880 

49 

12 

542906 

57936 

999997 

01 

642909 

57933 

4.57091 

48 

13 

577668 

53641 

999997 

01 

577672 

53642 

422328 

47 

14 

609853 

49933 

999996 

01 

609857 

49939 

390143 

46 

15 

639816 

46714 

999996 

01 

639820 

46715 

360180 

45 

16 

667845 

43881 

999995 

01 

667849 

43882 

332151 

44 

17 

694173 

41372 

999995 

01 

694179 

41373 

30582  i 

43 

18 

718997 

39135 

999994 

01 

719003 

39136 

280997 

42 

19 

742477 

3/127 

999993 

01 

742484 

37128 

2.57:- 16 

41 

20 

764754 

3.5315 

999993 

01 

764761 

J5136 

235239 

40 

21 

7.785943 

33672 

9. 99999 . 

01 

7.785951 

33673 

12.214U49 

39 

22 

806146 

32175 

999991 

01 

8061.55 

.32176 

19.3845 

38 

23 

82545J 
843931 

30805 

999990 

01 

825460 

30806 

174.540 

37 

24 

29547 

999989 

02 

843944 

29549 

1.56056 

36 

25 

861662 

28388 

999988 

02 

861674 

2S390 

138326 

35 

26 

878695 

27317 

999988 

02 

873708 

27318 

121292 

34 

27 

895085 

26323 

999987 

02 

895099 

26325 

104901 

33 

28 

910879 

25399 

999986 

02 

910894 

2.5401 

089106 

32 

29 

926119 

24538 

999935 

02 

926134 

24^^40 

073S66 

31 

30 

940842 

23733 

999933 

02 

940858 

23735 

059142 

30 

3f 

7.955082 

22980 

9.9990vS2 

02 

7.955100 

22981 

12.044J00 

29 

32 

968S70 

22273 

999981 

02 

968889 

22275 

031111 

28 

33 

982233 

21608 

9'Jl»  J  80 

02 

982253 

21610 

017747 

27 

b4 

995198 

20981 

9&  979 

02 

995219 

2  )983 

004781 

26 

35 

8.007787 

20390 

999977 

02 

8.007809 

2  1392 

11.992191 

25 

36 

020021 

19831 

999976 

02 

020045 

1)8:^3 

979955 

24 

37 

031919 

19302 

999975 

02 

031945 

19305 

968055 

23 

38 

04350 i 

18801 

999973 

02 

043527 

18803 

956473 

22 

39 

0.54781 

18325 

999972 

02 

054809 

18.327 

945 19 1 

21 

40 

065776 

17872 

999971 

02 

065806 

17874 

.  934194 

20 

41 

8.076.500 

17441 

9.iK)9969 

02 

8.076531 

17444 

11.923469 

19 

42 

086966 

17031 

999968 

02 

08699? 

17034 

913003 

18 

43 

097183 

16639 

999966 

02 

097217 

16042 

•902783 

17 

44 

107167 

16265 

999964 

03 

107202 

13268 

892797 

16 

45 

116926 

15908 

999963 

03 

116963 

15910 

883037 

15 

46 

•  126471 

1.5.566 

999961 

03 

126510 

l,5.56e 

873490 

!4 

47 

13.5810 

1.5233 

9999.59 

03 

13.5851 

1.5241 

864149 

13 

48 

144953 

14924 

999958 

03 

144996 

14^-^7 

855004 

12 

49 

1.53907 

14622 

999956 

03 

1.539.52 

14327 

846048 

11 

50 

162681 

14333 

9999.54 

03 

162727 

14336 

837273 

10 

51 

8.171280 

140.54 

9.  J9yu.52 

03 

8.171328 

1405'< 

11  828672 

9 

52 

179713 

13786 

9999.50 

03 

179763 

13790 

820237 

8 

53 

187985 

13.529 

999948 

03 

188036 

13532 

811964 

7 

54 

196102 

13280 

999946 

03 

1961.56 

13284 

,  803844 

6 

55 

204070 

13J41 

999944 

3 

204126 

13044 

79.5874 

5 

56 

211895 

12810 

999942 

4 

211953 

12814 

7880/.  7 

4 

67 

219581 

12.587 

999940 

04 

219:541 

12590 

780359 

3 

58 

227134 

12372 

999938 

01 

227195 

12.37F 

772805 

2 

60 

231557 

12164 

999936 

01 

23462 

12168 

765379 

1 

60 

241855 

11  F3 

999934  04 

24I9>I 

11)17 

758079 

0 

n 

rNHmT"! 

Sine    1 

(;..iu.jr.  1 

•IW   iM.| 

m  Degiecd. 


SIXES  AND  TANGENTS.  (1  Degree.] 

10 

A. 

Sinu   1 

l>. 

Cosine   1  1). 

r-.uiii. 

D 

<'()t;iim.   1 

6 

8.241855 

11963 

9.999^34 

04 

8.241921 

11967 

ll.V 580791  60 

I 

249033 

11768 

999932 

04 

249102 

11772 

75089^  y-y 

2 

256094 

11580 

999929 

04 

256185 

11584 

743835  .jH 

3 

263042 

11398 

999927 

04 

263115 

1 1402 

736885  57 

4 

269881 

11221 

999925 

04 

269956 

11225 

7300441  56 

5 

276814 

11050 

999922 

04 

276691 

11054 

7233091 55 

6 

2S3243 

10883 

999920 

04 

283323 

10S87 

716677 

54 

7 

289773 

10721 

999918 

04 

2898.50 

10728 

710144 

53 

8 

296207 

10565 

999915 

04 

296292 

10570 

703708 

52 

9 

302546 

10413 

999913 

04 

302634 

10418 

697366 

51 

10 
11 

308794 
8.314954 

10266 

999910 
9.999907 

04 
04 

308884 

10270 
10126 

691116 

50. 
49 

10122 

8.315048 

11.684954 

12 

321027 

9982 

999905 

04 

321122 

9987 

678878 

4.S 

13 

327016 

9847 

999902 

04 

327114 

9851 

672888 

47 

14 

332924 

9714 

999899 

05 

333025 

9719 

686975 

46 

15 

338753 

9586 

999897 

05 

338S56 

9590 

681144 

45 

IG 

344504 

9460 

999894 

05 

344610 

9465 

655390 

44 

17 

.350181 

9338 

999891 

05 

350289 

9343 

64971 1 

43 

18 

355783 

9219 

999888 

05 

355895 

9224 

644105 

42 

19 

361315 

9103 

999885 

05 

301430 

9108 

638570 

41 

20 

366777 

8990 

999882 

05 

368895 

8995 

633105 

40 

'Zl 

8.372171 

8880 

9.999879 

05 

8.372292 

8885 

11.627708 

39 

22 

377499 

8772 

999876 

05 

377622 

8777 

622378 

38 

23 

382762 

8667 

999873 

05 

382889 

8672 

617111 

37 

24 

337962 

8564 

999870 

05 

388092 

8570 

611908 

36 

25 

393101 

8464 

999867 

05 

393234 

8470 

606766 

35 

20 

.398179 

8366 

999864 

05 

398315 

8371 

601685 

34 

27 

403199 

8271 

999861 

05 

403338 

8276 

596662 

33 

28 

408161 

8177 

999858 

05 

408304 

8182 

691696 

32 

29 

41.3068 

8086 

999854 

05 

413213 

8091 

686787 

3i 

30 

417919 

7996 

999851 

06 

418068 

8002 

581932 

30 

31 

8.422717 

7909 

9.999848 

06 

8.422869 

7914 

11.577131 

29 

32 

427462 

7823 

999844 

06 

427618 

7830 

672332 

28 

33 

432156 

7740 

999841 

06 

432315 

7745 

667685 

27 

34 

436800 

7657 

999838 

06 

436962 

7663 

6630331  26  | 

35 

441394 

7577 

999834 

06 

441580 

7583 

658440 

25 

36 

445941 

7499 

999831 

06 

448110 

7505 

553890 

24 

37 

450440 

7422 

999827 

06 

450813 

7428 

549387 

23 

38 

454893 

7346 

999823 

06 

455070 

7352 

544930 

22 

39 

459301 

7273 

999820 

06 

459481 

7279 

640519 

21 

40 
41 

463665 
8.467985 

7200 

993816 
9.999812 

06 
06 

463849 
8.468172 

7206 

638151 

20 
19 

7129 

7135 

11.531828 

42 

472263 

7060 

999809 

06 

472454 

7066 

527.546 

18 

43 

473498 

6991 

999805 

06 

476693 

8993 

523307 

17 

44 

480693 

6924 

999801 

06 

480892 

6931 

51910S 

IG 

45 

484848 

6859 

999797 

07 

485050 

6865 

514950 

15 

46 

489963 

6794 

999793 

07 

439170 

8801 

610830 

14 

47 

493040 

6731 

999790 

07 

493250 

6738 

506750 

13 

48 

497078 

6869 

999788 

07 

497293 

6876 

502707 

12 

49 

501080 

6608 

999782 

07 

601298 

6616 

498702 

11 

50 

505045 

6548 

999778 

07 

505267 

8555 

49-1733 

10 

5l' 

8.508974 

6489 

9.999774 

07 

8.509200 

6496 

11.490300 

9 

52 

512867 

6431 

999769 

07 

513098 

6439 

48H902 

8 

53 

516726 

6375 

999765 

07 

516961 

6382 

483039 

7 

54 

520551 

6319 

999761 

07 

520790 

6326 

479210 

6 

55 

524343 

6264 

999757 

07 

524586 

6272 

475414 

5 

56 

.528102 

6211 

999753 

07 

523349 

6218 

471651 

4 

57 

531828 

6158 

999748 

07 

532080 

6165 

467920 

3 

58 

535523 

6108 

999744 

07 

535779 

6113 

464221 

2 

59 

539186 

6055 

999740 

07 

539447 

8062 

460553 

I 

CO 

.542819 

6004 

999 73^ 

07 

543;)8 ! 

6012 

456916 

0 

n 

Cosine 

i   Si.c   |. 

C'.;lan^. 

Tiiiii?    |M.  1 

tib  Degrees 


so 

(2  Derrn 

3es.)  A 

TAHLE  OF  LOGAKITHMIC 

M. 

.-'IPK 

1). 

«'<!siiie   1  l». 

'P.u. ir.  i   n. 

1  r.........  1 

0 

.•3..)'i:^>iy 

tkh)  1 

9. 999 735 

[07 

8..543J>S4|  6.)'-^ 

ll  i  .45(iJiOi  u.> 

I 

545422 

5955 

999731 

07 

.548691 

5962 

4533091 59 

2 

549J95 

5998 

999726 

07 

.550268 

5914 

449732 i  53 

3 

553539 

5858 

9^9722 

08 

553817 

5866 

448183;  .57 

4 

557054 

5811 

999717 

08 

557336 

5319 

442664'  58 

5 

5fiJ540 
583999 

5765 

999713 

08 

560828 

5773 

43917yL55 

6 

5719 

999708 

08 

56429  I 

5727 

435709;  54 

7 

507431 

5674 

999704 

08 

567727 

5682 

432273;  53 

8 

570S30 

5639 

999899 

08 

571137 

5638 

428383  52 

9 

574214 

5537 

999894 

08 

•  574520 

5595 

425480;  5! 

10 

5775613 

5544 

999689 

08 

577877 

5552 

422123:50 

IT 

S.bSOSdZ 

5592 

9.999635 

08 

8.581208 

5510 

11.418  792  49 

12 

584193 

5469 

999680 

03 

.584514 

5468 

415486143 

13 

6-37469 

5419 

999675 

08 

587795 

5427 

412205:47 

M 

599721 

5379 

999670 

03 

591051 

5337 

408919;  46 

15 

593948 

5339 

999865 

05 

594283 

5347 

405717:  45 

ffi 

597152 

5300 

999660 

03 

597492 

5308 

402508,  44 

17 

699332 

5261 

999855 

03 

600877 

5270 

399323;  43 

18 

693489 

5223 

999850 

08 

603839 

5232 

3981611  42 

19 

696623 

5186 

999645 

09 

606978 

5194 

3930221  41 

20 

699734 

5149 

999840 

09 

610094 

51.58 

339996;  40 

21 

8.612S23 

5112 

9.999635 

09 

8.613189 

5121 

11.338311  39 

22 

615891 

.5976 

999829 

09 

616262 

5035 

383733;  33 

23 

618937 

5941 

999824 

09 

619313 

.5050 

3306371  37 

24 

621962 

5998 

999819 

09 

622343 

5015 

377857!  36 

25 

624965 

4972 

999814 

ol; 

625352 

4931 

3746481  35 

2t) 

627948 

4938 

999803 

09 

628340 

4947 

3ri680;  34 

27 

639911 

4904 

999803 

09 

631398 

4913 

388692;  33 

2S 

633354 

4871 

999597 

09 

634256 

4880 

365744  32 

29 

63t)776 

4839 

999592 

09 

637184 

4848 

382316  31 

:v.) 

639689 

4806 

999586 

09 

640093 

4816 

359907  30 

31 

8.642563 

4775 

9.999531 

09 

8 . 642982 

4784 

11.357018;  29 

32 

645428 

4743 

999575 

09 

645353 

•  4753 

354147|  28 

31 

648274 

4712 

999570 

09 

648704 

4722 

35I296i  27 

31 

601102 

4632 

999564 

09 

651537 

4691 

34S463'  26 

35 

653911 

4652 

999558 

10 

654352 

4661 

345648|  25 

33 

656702 

4822 

999553 

10 

057149 

4631 

34285 1 1  24 

37 

659475 

4592 

999547 

10 

G5992S 

4692 

3400721  23 

3S 

662230 

4583 

999541 

10 

662639 

4573 

337311122 

39 

664968 

4535 

999535 

10 

665433 

4544 

334567121 

40 
41 

6G7639 
8.6r0393 

4506 
4479 

999529 
9.999524 

10 

To 

688160 
8.670870 

4526 

33 1 840 1  20 
11. 3291391  19 

4438 

42 

673039 

4451 

999518 

10 

673563 

4461 

3284371  18 

43 

675751 

4424 

999512 

10 

676239 

4434 

323761  17 

44 

678405 

4397 

999506 

10 

678900 

4417 

3211001  16 

45 

>  681043 

4370 

999590 

10 

681.544 

4380 

3184581  15 

46 

683665 

4344 

999493 

10 

684172 

4354 

31.58281  14 

47 

636272 

4318 

999487 

10 

686784 

4328 

313216!  13 

43 

633363 

4292 

999481 

10 

639381 

4303 

3106!9|  12 

49 

691433 

4267 

99)475 

10 

691963 

4277 

308037;  1 1 

5) 

69399^ 

4242 

999489 

10 

694529 

4252 

39547  ij  10 

51 

8.696543 

4217 

9.99r»4H3 

ll 

8.697081 

4228 

11.302919  9 

52 

699973 

4192 

999456 

11 

699617 

4203 

3003-53i  8 

53 

791589 

4168 

999450 

11 

702139 

4179 

29786 1 j  7 

54 

704099 

4144 

999443 

11 

704646 

4155 

'  295354!  6 

55 

706577 

4121 

999437 

11 

707140 

4132 

2928fH)|  5 

SS 

709949 

4097 

999431 

11 

709818 

4108 

29i>?32'  4 

57 

711507 

4074 

999424 

11 

712083 

4085 

237917  3 

58 

713952 

4051 

999418 

11 

714534 

4062 

235465;  2 

5!) 

7 1 63 -{3 

40^9 

999411 

11 

716972 

4040 

23:^028  j  1 

fiO 

_7l_SS00 

40;)6 

9'^M9l 

11 

719398 

4')  17 

28.n(»4'  0 

C-^tnti 

1 

Smh    1 

.;....-...t:.  j      1 

i:m.c.     pr 

UuKif 


•^; 


W 


SINF.3  AND  TANGE?fTS.   ^3  DcgVCCS.^ 

21 

0 

M..,.   1 

1).   1 

<;n.«irie   1  It  1 

T....    1    M 

<'..•,,.....   1 

8.718800 

4003 

9.99Ji'»4( 

11 

8.719396 

4017 

ll.280«)04i  hu 

1 

721204 

3984 

99939b 

11 

721806 

3995 

278l9l!.'i 

2 

723595 

3982 

999391 

11 

7242 J4 

?..»74 

275796  ..a 

3 

725972 

3941 

999384 

11 

726588 

3952 

2734  12  57 

4 

728337 

3919 

999378 

LI 

728959 

3930 

271041  .56 

5 

730(588 

3898 

999371 

11 

731317 

3909 

2686S3I  .55 

6 

733027 

3877 

999364 

12 

733663 

3889 

266337 1  54 

7 

735354 

3S57 

999357 

12 

735996 

3868 

;^848 

264004  53 

8 

7370o7 

3835 

999350 

12 

738317 

261683152 

9 

739969 

3816 

999343 

12 

740626 

3827 

2.59374151 

ill 

742259 

3796 

999336 

12 

742922 

3807 

25707  8 

50 

li 

8.744536 

3776 

9.999329 

12 

8.74.5207 

3787 

11.2.54793 

49 

12 

746802 

3756 

999322 

12 

747479 

3768 

252521 

48 

la 

749055 

3737 

999315 

12 

749740 

3749 

250260 

47 

14 

751297 

3717 

999308 

12 

751989 

3729 

248011 

46 

15 

753528 

3698 

999301 

12 

7.54227 

3710 

245773 

45 

IR 

755747 

3679 

999294 

12 

756453 

.3692 

243547 

44 

17 

757955 

3661 

999286 

12 

758668 

3673 

241332 

43 

18 

760151 

3642 

999279 

12 

760872 

36.55 

239128 

42 

ly 

762337 

3624 

999272 

12 

763965 

3636 

236935 

41 

20 

764511 

3606 

999265 

12 

765246 

3618 

234754 

4«) 

2i 

8.766675 

3588 

9.9992.57 

12 

8.767417 

3600 

11.232583 

39 

22 

768828 

3570 

9992.50 

13 

769578 

3583 

23{)422 

38 

23 

770970 

3553 

999242 

13 

771727  3.565 

228273 

37 

24 

773101 

3535 

999235 

13 

773866 

3548 

226134 

36 

25 

775223 

3518 

999227 

13 

775995 

3531 

224005 

35 

2fi 

777333 

3501 

999220 

13 

778114 

3514 

22188H 

34 

27 

779434 

3484 

999212 

13 

780222 

3497 

219778 

33 

28 

781.524 

3467 

999205 

13 

782320 

34S0 

217680 

32 

29 

783605 

3451 

999197 

13 

784408 

3464 

21.5,592 

31 

:H) 

785675 

.3431 

999189 

13 

786486 

344  7 

213514 

30 

31 

8.787736 

3418 

9.999181 

13 

8.788.554 

3131" 

11.211446 

29 

32 

789787 

3402 

999174 

13 

790613 

3414 

209387 

28 

33 

791828 

3386 

999166 

[3 

792662 

3399 

207338 

27 

3t 

793859 

3370 

9991.5S 

13 

7947(»l 

3383 

205299 

26 

35 

795881 

3354 

999150 

13 

796731 

3368 

203269 

25 

3f) 

797894 

3339 

999142 

13 

79S752 

3352 

201248 

24 

37 

"799897 

3323 

999134 

13 

800763 

3337 

199237 

23 

3S 

801892 

3398 

999126 

13 

802765 

3322 

197235 

22 

39 

803876 

3293 

999118 

13 

804758 

3307 

195242 

21 

40 

805852 

3278 

999110 

13 

80674 2 1  3292 

1932.58 

20 

41 

8.807819 

3263 

9.999102 

13 

8.808717  3278 

11.191283 

19 

42 

809777 

3249 

999094 

14 

810683  3262 

189317 

18 

43 

811726 

3234 

9990861  14 

812641  3248 

187359 

17 

44 

813667 

3219 

9990771 14 

814589  3233 

18.5411 

16 

45 

815599 

3205 

999069 

14 

816529 

3219 

183471 

15 

46 

817522 

3191 

999061 

14 

818461 

3205 

181.539 

14 

47 

819436 

3177 

999053 

14 

820384 

3191 

179616 

13 

48 

821343 

3163 

999044 

14 

822298 

3177 

177702 

12 

49 

823240 

3149 

999036 

14 

824205 i  3163 

175795 

11 

50 

8251.30 

31.35 

999027 
9.999019 

14 
14 

826103 
8.827992 

3150 
3f36 

173897  10  1 
11.172008  9  1 

8.827011 

3122 

52 

828884 

3108 

999010 

14 

829874 

3123 

170126 

8 

53 

830749 

3095 

999002 

14 

831748 

3110 

1682.52 

7 

54 

832607 

3082 

998993 

14 

833613 

3096 

166.387 

6 

55 

834456 

3069 

998984 

14 

835471 

3083 

164.529 

5 

5fi 

83'^297 

3056 

998976 

14 

837321 

3070 

162679 

4 

57 

838130 

3043 

998967 

15 

8J9163 

3057 

160837 

3 

58 

839956 

3030 

998958 

15 

840998 

3045 

1590021  2 
1.57175   1 

59 

841774 

3017 

99S950 

15 

842825 

3032 

no 

843585 

3000 

9989H 

15 

8446441  3019 

1.553561  0 

z 

(J<i.-ine 

1 

Si,e    j 

Vv.vms.     1 

Ta  .!■.    \M 

86  Uuifrcus. 


22 

( 

4  Degrees.')  a 

TABLE  OF  LOGAKlTir.MlC 

'.VI 

f^iiif 

1   n 

Cosine   1  1). 

_:!:^^_ 

__!'_ 

(Manj:. 

"(T 

8. y 43585 

3005 

J. 99894 11  15 

8.844644 

3019 

ii.  l5.5:}5Tr 

fiT 

1 

845387 

2992 

998932  15 

846455 

3007 

153545 

"59 

o 

847183 

2980 

998923  15 

848260 

2995 

151740 

58 

3 

848971 

2967 

998914  15 

850057 

2982 

149943 

57 

4 

8507Jil 

2955 

998905  15 

851846 

2970 

148154 

56 

5 

852525 

2943 

998896  15 

853628 

2958 

146372 

55 

i\ 

854291 

2931 

998887  15 

8.5.5403 

2946 

14459/' 

54 

7 

856049 

2919 

998378  15 

857171 

2935 

142829 

53 

8 

857801 

2907 

998869  15 

858932 

2923 

141068 

52 

9 

859546 

289e 

998860  15 

860686 

29  1 1 

139314 

51 

10 

861283 

2884 

998851  15 

862433 

2900 

137567 

50 

11 

8.863014 

~2873 

9.998841 

15 

8.864173 

2888 

11.135827 

49 

12 

864738 

2861 

993832 

15 

865906 

2877 

134094 

48 

13 

866455 

2850 

Q93823 

16 

867632 

2866 

132368 

47 

14 

868165 

2339 

993813 

16 

869351 

2854 

130649 

46 

15 

869368 

2828 

998804 

16 

871064 

2843 

128936 

45 

IT) 

871565 

2817 

998795 

16 

872770 

2832 

127230 

44 

17 

873255 

2S06 

998785 

16 

874469 

2821 

125.531 

43 

18 

874938 

2795 

998776 

16 

876162 

2811 

123833 

42 

19 

876615 

2786 

998766 

16 

877849 

2800 

122151 

41 

20 

21 

878285 
8.87994'J 

2773 

998757 
9.993747 

16 
16 

879529 
8.881202 

2789 

120471 

40 
39 

2763 

2779 

11.118798 

22 

881607 

2752 

998733 

16 

882S69 

2788 

117131 

38 

23 

883258 

2742 

998728 

16 

884530 

2758 

11.5470 

37 

24 

884903 

2731 

998718 

16 

886185 

2747 

113815 

3f) 

25 

886542 

2721 

998708 

16 

887833 

2737 

112167 

35 

20 

888174 

2711 

993699 

16 

889476 

2727 

110524 

34 

27 

889801 

2700 

998689 

16 

891112 

2717 

108388 

33 

28 

891421 

2690 

998679 

16 

892742 

2707 

107258 

.32 

29 

893035 

2680 

998669 

17 

894ii66 

2697 

105634 

31 

30 

894643 

2670 

998659 

17 

895984 

2687 

104016 

30 

31 

8.896246 

2660 

9.998649 

17 

8.897.596 

2677 

11.102404 

29 

32 

897842 

2651 

998639 

17 

899203 

2667 

10(»797 

28 

33 

899432 

2641 

998629 

17 

900803 

26.58 

099197 

27 

34 

901017 

2631 

998619 

17 

902398 

2648 

097602 

26 

35 

902596 

2622 

998609 

17 

903987 

2638 

096013 

25 

36 

904169 

2612 

998599 

17 

905570 

2629 

094430 

24 

37 

905736 

2603 

998589 

17 

907147 

2620 

0928.53 

23 

38 

907297 

2593 

998578 

17 

908719 

2610 

091281 

22 

39 

908353 

2584 

998568 

17 

910285 

2601 

039715 

21 

40 

910404 

2575 

998558 

17 

911846 

2592 

088154 

20 

41 

8.911949 

2560 

9.998.548 

17 

8.913401 

2583 

11.086599 

19 

42 

913488 

2150 

998.537 

17 

914951 

2574 

08,5049 

18 

43 

9150'^2 

2547 

998.527 

17 

916495 

2565 

083505 

17 

44 

916.55« 

2538 

998516 

18 

918034 

2556 

081966 

16 

45 

,  918073 

2529 

998.506 

18 

919568 

2547 

080432 

15 

46 

919591 

2520 

998495 

18 

921096 

2538 

078904 

14 

47 

921103 

2512 

998485 

18 

922619 

2530 

07738 1 

13 

48 

922610 

2.503 

998474 

18 

924136 

2.521 

075864 

12 

49 

924112 

2494 

998464 

18 

925649 

2512 

074351 

H 

50 

925609 

2486 

998453 

18 

927156 

2503 

072844 

10 

51 

8.927100 

24^7 

9.998442 

18 

8.9286.58 

2495 

11.071342 

9 

62 

928.587 

2469 

998431 

18 

9301.55 

2486 

069845 

8 

53 

930068 

2460 

998421 

18 

931647 

2478 

068353 

7 

54 

931.544 

2452 

998410 

18 

933134 

2470 

'  066H66 

6 

55 

933015 

2443 

908399 

18 

934616 

2461 

065384 

5 

56 

934481 

2435 

998388 

18 

936093 

2453 

063907 

4 

57 

935942 

2427 

998377 

18 

937565 

2445 

062435 

3 

58 

937398 

2419 

998366 

18 

9390.32 

2437 

060968 

2 

59 

93S850 

2411 

998355 

18 

940494 

2430 

059506 

1 

_60_ 

940296 

2403 

998344 

18 

941952 

2421 

058048 

0 

Co..;,e 

ijiii<>   1 

Ct>liin<! 

Tana. 

"mT' 

eS  DeKreen. 


SINKS  AND  TANGENTS.     (5  Degrees.) 


23 


M 

t^iiio 

I.. 

Cosine   j  1) 

i   iViMu.   1   I). 

1   C.au    ! 

^ 

6.i)^>UJ6 

■4iJ.i 

9 .  90  -id  k^ 

19 

8.91:19)* 

ZhZi 

1 1 .0.^dsIM]  60 

1 

94l7tJS 

2394 

998333 

10 

943404 

2413 

056596' 59 

2 

943174 

2387 

993822 

19 

944S52 

2405 

055148158 

a 

944606 

2379 

90 S3 11 

19 

546205 
^4773 1 

2397 

0.53705,  57 

4 

946034 

2371 

903300 

19 

2390 

052266  .5G 

f, 

947456 

2363 

993239 

19 

949168 

2382 

050832  55 

f. 

948 S74 

2355 

993277 

19 

950597 

2374 

049403  54 

V 

950287 

2343 

993266 

19 

952021 

2366 

047979  53 

b 

951696 

2340 

998255 

19 

95344 1 

2360 

046559!  52 

i 

953100 

2332 

998243 

19 

954856 

2351 

045144151 
0437331 50 

10 

954499 

2325 

998232 

19 

956267 

2344 

il' 

8.935894 

2317 

9.99322.) 

19 

8.957674 

2337 

11.0423:iGJ4ji 

Ik 

957284 

2310 

998209 

19 

959075 

2329 

040925  48 

i; 

958670 

2302 

993197 

19 

960473 

2323 

039527  47 

14 

980052 

2295 

903186 

19 

961866 

2314 

038134  46 

U 

961429 

2283 

993174 

19 

963255 

2307 

036745  45 

U> 

962801 

2280 

998163 

19 

964639 

2300 

035361  44 

17 

964170 

2273 

908151 

19 

966019 

2293 

033981  43 

IH 

965534 

2206 

998139 

20 

917394 

2286 

032606  42 

l!» 

966893 

2259 

998128 

20 

963768 

2279 

031234  41 

2;) 

96 -{249 

2252 

998116 

20 

970133 

2271 

029867  40 

2l" 

8.96080) 

2244 

9.998104 

20 

8.971496 

2265 

11.028504  39 

22 

970947 

2238 

908092 

20 

972855 

2257 

027145  33 

23 

972289 

2231 

998080 

20 

974209 

2251 

025791  37 

24 

973628 

2224 

998068 

20 

975560 

224-4 

0244401  36 

25 

974962 

2217 

998056 

20 

976906 

2237 

023094  35 

26 

978293 

2210 

993044 

20 

978248 

2230 

021752  34 

27 

977619 

2203 

998032 

20 

979588 

2223 

020414  33 

2S 

978941 

2197 

993020 

20 

9309211  2217 

019079  32 

29 

980259 

2190 

998008 

20 

932251 

2210 

0177491 31 

3f) 

9^1573 

2183 

997996 

20 

983577 

2204 

016423(30 

31 

8.982883 

2177 

9.997984 

20 

8.984899 

2197 

11.015101 

29 

3-7 

984189 

2170 

997972 

20 

936217 

2191 

013783 

28 

33 

985491 

2163 

997959 

20 

937532 

2184 

012468 

27 

3t 

986739 

2157 

997947 

20 

938842 

2178 

011158 

26 

35 

983083 

2150 

997935 

21 

990149 

2171 

009851 

25 

3^ 

989374 

2144 

997922 

21 

991-451 

2165 

008549 

24 

37 

900660 

2138 

997910 

21 

992759 

2158 

007250 

23 

3S 

931943 

2131 

9^7897 

21 

994045 

2152 

005955 

22 

33 

903222 

2125 

997885 

21 

995337 

2146 

004663 

21 

40 

901497 

.  2119 

907872 

21 

996624 

2140 

003376 

20 

41 

8.0J5768 

2112 

9.907860 

21 

8.997903 

2134 

11.002002 

19 

42 

907036 

2105 

997847 

21 

999188 

2127 

000812 

18 

43 

908290 

2100 

997335 

21 

9.000465 

2121 

10.999535 

17 

44 

999560 

2094 

937822 

21 

001738 

2115 

993262 

16 

45 

9. 000816 

2087 

997800 

21 

093007 

2109 

996993  15 

46 

002069 

2082 

907797 

21 

004272 

2103 

9057281 14 

47 

003318 

2076 

99778 1  21 

005534 

2097 

9944661 13 

48 

004563 

2070 

997771  21 

008792 

2091 

903208  12 

49 

005805 

2064 

997758  21 

003047 

2085 

991953  11 

50 

007044 

2058 

907745  21 

009298 

2080 

990702  10 

51 

9.00S278 

2052 

9.997732 

21 

9.010546 

2074 

10.989454;  9 

52 

009510 

2046 

907719 

21 

011790 

2068 

988210  6 

53 

010737 

2040 

907706 

21 

01 3031 

2062 

986969;  7 

54 

011953 

2134 

907693 

22 

014263 

2056 

985732  n 

55 

013182 

2029 

907680 

22, 

015502 

2051 

98 44 OS  5 

56 

014400 

2023 

9976671 22 

oituz-:: 

2045 

983268  4 

57 

0:5613 

2017 

907654  22 

017959 

2040 

982041  3 

5S 

016824 

2012 

907641  22 

019183 

2033 

9><fi8|7  2 

59 

0180311 

20)6 

907628  22 

020 1')3 

2028 

979597   1 

m 

oior.ri 

20  )) 

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0>i'»«"jO'  2023  1 

fl79'^90'  0 

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C  .>hi.;   1 

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Siue    1    1 

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1  MIL'    1  M 

17 

S4Ue 

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24 

(6  Depreps.)   a  • 

FAULK  OF  LOGAKITIIMIC     j 

1 

1 

M  j 

s.....   1   h. 

(  ii.-iiic   1  h 

•Innfi.   1   I).   1 

L'utf\ii<i.       i    1 

0 

9.iny:i35  XOOi) 

9.997614 

22 

9.02)620  2023 

10.978380 

Ji) 

I 

020435  1995 

997(i01 

22  1 

0228341  2017 

977166 

59 

2 

0216:{2  1989 

997.588 

22 

024044  2011 

975956 

58 

3 

022>i25'  1984 

99^574 
99^)61 

22 

025251  2006 

974741;' 

57 

4 

024016  1978 

22 

026455  2000 

973.545 

56 

S 

025203  1973 

997.547 

22 

0276551  1995 

972345 

55 

6 

026386  1967 

997.5.34 

23 

028852!  1990 

971  14 H 

54 

7 

0275671  1962 

997520 

23 

03004 6 1  1985 

969954 

53 

8 

(»28744;  1957 

997607 

23 

0312371  1979 

96876;] 

52 

9 

029*}  18  1951 

997493 

23 

032425!  1974 

967575 

51 

10 

031089  1947 

9974S0 

23 

0.336091  1969 

f  66.391 

50 

ll 

9.032257;  1941 

9.997466 

23 

9. 03479 ll  1964 

I0.965'>d9 

49 

12 

03342  ll  1936 

9974.52 

23 

0359691  19.58 

964031 

iS 

13 

0345821  1930 

997139 

23 

037144  1953 

962856 

47 

14 

0357411  1925 

997425 

23 

038316  1948 

961684 

46 

15 

036896!  1920 

997411 

23 

0394851  1943 

960515 

45 

16 

038048 

1915 

997397 

23 

040651   1938 

959349 

44 

17 

039197 

1910 

997383 

23 

041813!  1933 

9.58187 

43 

18 

040342 

1905 

997369 

23 

042973  1928 

957027 

42 

19 

041485 

1899 

997355 

23 

0441301  1923 

955870 

41 

20 

042625 

1894 

997341 

23 

04.52841  1918 

9.54716 

40 

21 

9. 0437621  1889 

9.997327 

24 

9.0464341  1913 

10.9.53566 

39 

22 

0448951  1884 

997313 

24 

047582 

1908 

952418 

38 

23 

046026!  1879 

997299 

24 

048727 

1903 

95127:^ 

37 

24 

047154  1875 

997285 

24 

049869 

1898 

950131 

36 

25 

048279  1870 

997271 

24 

051008   1893 

948991 

35 

2G 

049400  1865 

997257 

24 

0,52144  1889 

947866 

34 

27 

050519  1860 

997242 

24 

053277  1884 

94672S 

33 

28 

051635 

1855 

997228 

24 

054407  1879 

945593 

32 

29 

052749 

1850 

997214 

24 

05.55351  1874 

944466 

31 

30 

053859 

1845 

997199 

24 

0.56659  1870 

943.341 

30 

nl 

0549661  1841 

9.99718.5 

24 

9.0.57781   1865 

10.942219 

29 

32 

0560711  1836 

997170 

24 

058900  1869 

94  HOC 

28 

33 

057172 

1831 

997156 

24 

060016  18.55 

939984 

27 

34 

058271 

1827 

997141 

24 

0611.30  1851 

938870 

26 

35 

0593G7 

1822 

997127 

24 

0G2240  1846 

9.37760 

25 

36 

060460 

1817 

997112 

24 

063348  1842 

9.3665i 

24 

37 

061551 

1813 

997098 

24 

0644531  1837 

935.547 

23 

38 

062639 

1808 

997083 

25 

065556 

1833 

934444 

22 

39 

063724 

1804 

997068 

25 

0666.55 

1828 

93334.': 

21 

40 

064806 

1799 

9970.53 

25 

067752 

1824 

935^248 

20 

41 

9.0658851  1794 

9.9970.39 

25 

9.068846 

1819 

10.93  fl. 54 

19 

42 

066962 

1790 

997024 

25 

0099.38 

1815 

93U06i 

18 

43 

068036 

1786 

997009 

25 

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SINES  A^'D  TAKCFNTS.  {^9   Degrees. 

; 

27 

M. 

Sine   1   D. 

Cosinfi   1  D 

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«. 

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Ol 9.194332 

1328 

9.994620 

33 

9.199713 

1361 

10.800287;  60 

1 

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1326 

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33 

200.529 

1359 

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2 

195925 

1323 

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59 

3 

196719 

1321 

904560 

34 

202159 

1354 

797841 

57 

4 

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1318 

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34 

202971 

1352 

797029 

66 

5 

198302 

1316 

994519 

34 

203782 

1349 

790218 

56 

6 

199091 

1313 

994499 

34 

204592 

1347 

705408 

54 

7 

199879 

1311 

994479 

34 

205400 

1345 

794600 

53 

8 

200666 

1308 

994459 

34 

206207 

1342 

793793 

52 

9 

201451 

1306 

994438 

34 

207013 

1340 

792987 

51 

10 
11 

202234 

1304 
1301 

994418 
9.994397 

34 
34 

207817 

1338 
1335 

792183 
10.791381 

60 
49 

9.203017 

9.208619 

12 

203797 

1299 

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34 

209420 

1333 

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48 

13 

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1296 

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34 

210220 

1331 

789780 

47 

14 

205354 

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34 

211018 

1328 

788982 

46 

15 

206181 

1292 

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34 

211815 

1326 

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45 

16 

206906 

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34 

212611 

1324 

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44 

17 

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35 

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18 

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19 

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214989 

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30 
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35 
35 

215780 
9.216568 

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40 
39 

9.210760 

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22 

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24 

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25 

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35 

26 

214579 

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35 

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1301 

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34 

27 

215338 

1264 

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35 

221272 

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33 

28 

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29 

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31 

30 
31 

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35 
35 

223606 

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30 
29 

9.218363 

9.993981 

9.224382 

1290 

10.775618 

32 

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33 

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27 

34 

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26 

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36 

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1281 

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25 

36 

222115 

1244 

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36 

228239 

1279 

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24 

37 

222861 

1242 

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36 

229007 

1277 

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23 

38 

223606 

1239 

993832 

36 

229773 

1275 

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22 

39 

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36 

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1273 

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21 

40 
41 

225092 
9.225S33 

1235 
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36 

36 

231302 

1271 

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20 
19 

9.993768 

9.232065 

1269 

10.767935 

42 

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36 

232826 

1267 

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18 

43 

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1228 

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36 

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17 

44 

228048 

1220 

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36 

234345 

1262 

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16 

45 

228784 

1224 

993681 

36 

235103 

1260 

764897 

15 

46 

229518 

1222 

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38 

235859 

1258 

764141 

14 

47 

230252 

1220 

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36 

236614 

1256 

763386 

13 

48 

230984 

1218 

993616 

36 

237368 

1254 

762632 

12 

49 

231714 

1216 

993594 

37 

238120 

1252 

761880 

11 

50 
61 

232444 
9.233172 

1214 
1212 

993572 
9.99.3550 

37 
37 

238872 

1250 

761128 
10.760378 

io 

9 

9.239622 

1248 

52 

233899 

1209 

993528 

37 

240371 

1246 

759620 

8 

53 

234625 

1207 

993.506 

37 

241118 

1244 

758882 

7 

54 

235349 

1205 

993484 

37 

241865 

1242 

758135 

6 

55 

236073 

1203 

993462 

37 

242610 

1240 

757390 

5 

56 

236/95 

1201 

993440 

37 

243354 

1238 

756646 

4 

57 

237515 

1199 

993418 

37 

244097 

1236 

755903 

3 

58 

238235 

1197 

993396 

37 

244839 

1234 

7.55161 

2 

59 

238958 

1195 

993374 

37 

245579 

1232 

764421 

1 

60 

239670 

1193 

993351 

37 

246319 

1230 

753681 

0 

i   1 

Cosine   i       i 

Siae   1   1 

Cotaiig.  1 

Tang.   |M.  1 

17* 


8U 


EE 


28 

(10  Dogr 

ees.)  A 

TABLE  OF  LOGARITHMIC 

M. 

Situ; 

D. 

vvosine   1  D. 

Tnus. 

D. 

1   Cotano.  1   1 

"o" 

9.239670 

1193 

9.993351 

37 

9.246319 

1230 

10.753081 

60 

1 

240386 

1191 

993329 

37 

247057 

1228 

752943 

69 

2 

241101 

1189 

993307 

37 

247794 

1226 

762206 

58 

3 

241814 

1187 

993286 

37 

248530 

1224 

761470 

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4 

242526 

1186 

993202 

37 

249264 

1222 

750736 

66 

6 

243237 

1183 

993240 

37 

249998 

1220 

750002 

55 

6 

243947 

1181 

993217 

38 

260730 

1218 

749270 

64 

7 

244656 

1179 

993195 

38 

261461 

1217 

748539 

63 

8 

245363 

1177 

993172 

38 

252191 

1215 

7478091 52  1 

9 

246069 

1176 

993149 

38 

252920 

1213 

7470801 51 1 

10 
11 

246775 
9.247478 

1173 
1171 

993127 

38 
38 

263648 
9.264374 

1211 

7463.52 

50 
49 

9.993104 

1209 

10.745626 

12 

248181 

1169 

993081 

38 

256100 

1207 

744900 

48 

13 

248883 

l.o7 

993069 

38 

255824 

1205 

744176 

47 

14 

249583 

1165 

993036 

38 

266.547 

1203 

743453 

46 

15 

250282 

1163 

993013 

38 

257269 

1201 

742731 

45 

16 

250980 

1161 

992990 

38 

257990 

1200 

742010 

44 

17 

251677 

1159 

992967 

38 

258710 

1198 

741290 

43 

18 

252373 

1158 

992944 

38 

269429 

1196 

740571 

42 

19 

253067 

1156 

992921 

38 

260146 

1194 

7398.54 

41 

20 
21 

253761 

1164 
11,52 

992898 
9.992875 

38 
38 

260863 

1192 
1190 

739137 

4_0 
39 

9.254453 

9.261678 

10.738422 

22 

265144 

1150 

992852 

38 

262292 

1189 

737708 

38 

23 

255834 

1148 

992829 

39 

263005 

1187 

736995 

37 

24 

266523 

1146 

992806 

39 

263717 

1185 

736283 

36 

26 

257211 

1144 

992783 

39 

264428 

1183 

735572 

36 

26 

257898 

1142 

992759 

39 

266138 

1181 

734862 

34 

27 

268583 

1141 

992736 

39 

266847 

1179 

7341.53 

33 

28 

269268 

1139 

992713 

39 

266565 

1178 

733445 

32 

29 

259951 

1137 

992690 

39 

267261 

1176 

732739 

31 

30 
31 

260633 

3135 

992666 
9.992643 

39 
39 

267967 

1174 
1172 

732033 
10.731329 

30 

29 

9.261314 

1133 

9.268671 

32 

261994 

1131 

992619 

39 

269376 

1170 

730626 

28 

33 

262673 

1130 

992596 

39 

270077 

1169 

729923 

27 

34 

263351 

1128 

992572 

39 

270779 

1167 

729221 

26 

35 

264027 

1126 

992649 

39 

271479 

1165 

728621 

25 

36 

264703 

1124 

G92626 

39 

272178 

1164 

727822 

24 

37 

265377 

1122 

992601 

39 

272876 

1162 

727124 

23 

38 

266051 

1120 

992478 

40 

273573 

1160 

726427 

22 

39 

266723 

1119 

992464 

40 

274269 

1158 

725731 

21 

40 

267395 

1117 

992430 

40 

274964 

1157 

725036 

20 

41 

9.268065 

1115 

9.992406 

40 

9.276668 

1156 

10.724342 

19 

42 

268734 

1113 

992382 

40 

276361 

1153 

723649 

18 

43 

269402 

1111 

992359 

40 

277043 

1151 

722957 

17 

44 

270069 

1110 

992336 

40 

277734 

1150 

722266 

W5 

45 

270736 

1108 

992311 

40 

278424 

1148 

721576 

16 

46 

271400 

1106 

992287 

40 

279113 

1147 

720887 

14 

47 

272064 

1105 

992263 

40 

279801 

1146 

720199 

13 

48 

272726 

1103 

992239 

40 

280488 

1143 

719512 

12 

49 

273388 

1101 

992214 

40 

281174 

1141 

718826 

11 

60 
51 

274049 
9.274708 

1099 

992190 

40 
40 

281858 

1140 
1138 

718142 
10.717468 

50 
9 

1098 

9.992166 

9.282542 

52 

275367 

1096 

992142 

40 

283225 

1136 

716776 

8 

53 

276024 

1094 

992117 

41 

283907 

1136 

716093 

7 

54 

276681 

1092 

992093 

41 

284688 

1133 

,  715412 

6 

-55 

277337 

1091 

992069 

41 

286268 

1131 

714732 

5 

56 

277991 

1089 

992044 

41 

285947 

1130 

714063 

4 

57 

278644 

1087 

992020 

41 

286624 

1128 

713.376 

3 

58 

279297 

1086 

991996 

41 

287301 

1126 

712699 

2 

59 

279948 

1084 

991971 

41 

287977 

1125 

712023 

1 

60 

280599 

1082 

991947 

41 

288662 

1123 

711348 

0 

Cosine  1 

Bine 



Ctang.  1 

1 

Tang.   IM.| 

97  Degrees. 


siKES  AND  TANGENTS.    (11  Degrees.) 


29 


M. 

1   Sine 

1   n. 

1  Cosiiifi  1  n. 

i  Tgn.. 

1    D. 

1   Ccrta.,!:.   1  \ 

"(T 

[9.280599 

1082 

9.991947 

41 

9.288652 

1123 

10.711348,  60 1 

1 

281248 

1081 

991922 

41 

289326 

1122 

710674 

59 

2 

281897 

1079 

991897 

41 

289999 

112U 

710001 

58 

3 

282544 

1077 

991873 

41 

290671 

1118 

709329 

57 

4 

283190 

1076 

991848 

41 

291342 

1117 

708658 

56 

5 

283836 

1074 

991823 

41 

292013 

1115 

707987 

55 

6 

284480 

1072 

991799 

41 

292682 

1114 

707318 

54 

7 

285124 

1071 

991774 

42 

293350 

1112 

7066501531 

8 

285766 

1069 

991749 

42 

294017 

1111 

705983!  521 

9 

286408 

1067 

991724 

42 

294684 

1109 

705316 

51 

10 
11 

287048 
9.287687 

1066 

991699 
9.991674 

42 
42 

295349 

1107 

704651 

50 
49 

1064 

9.296013 

1106 

10.703987 

12 

288326 

1063 

991649 

42 

296677 

1104 

703323 

48 

13 

288964 

1061 

991624 

42 

297339 

1103 

702661 

47 

14 

289600 

1059 

991599 

42 

298001 

1101 

701999 

46 

15 

290236 

1058 

991574 

42 

208662 

1100 

701338 

45 

16 

290870 

1056 

991549 

42 

299322 

1098 

700678 

44 

17 

291.504 

1054 

991524 

42 

299980 

1096 

700020 

43 

IS 

293137 

1053 

991498 

42 

300638 

1095 

699362 

42 

19 

292768 

1051 

991473 

42 

301295 

1093 

698705 

41 

20 

293399 

1050 

991448 

42 

301951 

1092 

698049 

40 

21 

9.294029 

1048 

9.991422 

42 

9.302607 

1090 

10.697393 

39 

22 

294658 

1046 

991397 

42 

303261 

1089 

696739 

38 

23 

295286 

1045 

991372 

43 

303914 

1087 

696086 

37 

24 

295913 

1043 

991346 

43 

.304567 

1086 

695433 

36 

25 

296539 

1042 

991321 

43 

305218 

1084 

694782 

35 

26 

297164 

1040 

991295 

43 

305869 

1083 

694131 

34 

27 

297788 

1039 

991270 

43 

306519 

1081 

693481 

33 

28 

298412 

1037 

991244 

43 

307168 

1080 

692832 

32 

2 'J 

299034 

1036 

991218 

43 

307815 

1078 

692185 

31 

3!) 
31 

299655 
9.300276 

1034 

991193 
9.991167 

43 
43 

308463 
9.309109 

1077 

691537 

30 
29 

1032 

1075, 

10.690891 

32 

300895 

1031 

991141 

43 

309754 

1074 

690246 

28 

33 

301514 

1029 

991115 

43 

310398 

1073 

689602 

27 

34 

302132 

1028 

991090 

43 

811042 

1071 

688958 

26 

35 

302748 

1026 

991064 

43 

311685 

1070 

688315 

25 

36 

303364 

1025 

991038 

43 

312327 

1068 

687673 

24 

37 

303979 

1023 

991012 

43 

312967. 

1067 

687033 

23 

38 

304593 

1022 

990986 

43 

313608 

1065 

686392 

22 

39 

305207 

1020 

990960 

43 

314247 

1064 

685753 

21 

40 

305319 

1019 

990934 

44 

314885 

1062 

685115 

20 

41 

9.306430 

1017 

9.990908 

44 

9.315523 

1061 

10.684477 

19 

42 

307041 

1016 

990882 

44 

316159 

1060 

683841 

18 

43 

307650 

1014 

990855 

44 

316795 

1058 

683205 

17 

44 

308259 

1013 

990829 

44 

317430 

1057 

682570 

16 

45 

308887 

1011 

990803 

44 

318064 

1055 

681936 

15 

46 

309474 

1010 

990777 

44 

318697 

1054 

681303 

14 

47 

310080 

1008 

990750 

44 

319329 

1053 

680671 

13 

48 

310685 

1007 

990724 

44 

319961 

1051 

680033 

12 

49 

311289 

1005 

990697 

44 

320592 

1050 

679408 

11 

50 

311893 

1004 

990671 

44 

321222 

1048 

678778 

10 

51 

9.312495 

1003 

9.990644 

44 

9.321851 

1047 

10.678149 

9 

52 

313097 

1001 

990618 

44 

322479 

1045 

677521 

8 

53 

313698 

1000 

990591 

44 

323106 

1044 

676894 

7 

54 

314297 

998 

990565  44 

323733 

1043 

676267 

6 

55 

314897 

997 

990538  44 

324358 

1041 

675642 

5 

56 

315495 

996 

990511  45 

324983 

1040 

675017 

4 

57 

316092 

994 

J90485  45 

325607 

1039 

674393  3 

58 

316689 

993 

990458  45 

326231 

1037 

673769  2 

59 

317284 

991 

990431  45 

326853 

1036 

673117  1 

€0 

317879 

990 

990404  45 

327475 

10.35 

672525  0 

I   Cosine 

•   8iiie   1 

)  (Jolaiu. 

Tan-,   j 

78  Degrees 


30 

(12  Degrees.)  a 

TABLE  OP  I.OGARITHMIC 

M. 

1    Si.ie 

!    i>. 

1   Cosine   |  D. 

1  Timti. 

1   D. 

1   Cofaii;;.  i   1 

"IT 

9.317879 

990 

9.&9U404 

45 

9.327474 

1035 

10.672526 

60 

1 

318473 

988 

990378 

45 

328095 

1033 

671905 

59 

2 

319066 

987 

990351 

45 

328715 

1032 

671285 

58 

3 

319658 

986 

990324 

45 

329334 

1030 

670G66 

57 

4 

320249 

984 

990297 

45 

329953 

1029 

670047 

56 

5 
6 

320840 

983 

990270 

45 

330570 

1028 

669430 

55 

"3til475tr 

~  982 

990243 

45 

331187 

1026 

668813 

54 

7 

322019 

980 

990215 

45 

331803 

1025 

668197 

53 

8 

322607 

979 

990188 

45 

332418 

1024 

667582 

52 

9 

323194 

977 

990161 

45 

333033 

1023 

066967 

51 

10 
11 

323780 
9.324366 

976 

990134 
9.990107 

45 

46 

333646 

1021 

666354 
10.66.5741 

50 
49 

975 

9.334259 

1020 

12 

324950 

973 

990079 

46 

334871 

1019 

665129 

48 

13 

325534 

972 

990052 

46 

335482 

1017 

664518 

47 

14 

326117 

970 

990025 

46 

336093 

1016 

663907 

46 

15 

326700 

969 

98'.997 

46 

336702 

1015 

663298 

45 

16 

327281 

968 

989970 

46 

337311 

1013 

662689 

44 

17 

327862 

966 

989942 

46 

337919 

1012 

662081 

43 

18 

328442 

965 

989915 

46 

338527 

1011 

661473 

42 

19 

329021 

964 

989887 

46 

339133 

1010 

660867 

41 

20 
21 

329599 

962 

989860 

46 
46 

339739 

1008 
1007 

660261 
10.659656 

40 
39 

9.330176 

961 

9.989832 

9.340344 

22 

330753 

960 

989804 

46 

340948 

1006 

6590.52 

38 

23 

331.329 

958 

989777 

46 

341552 

1004 

658448 

37 

24 

331903 

957 

989749 

47 

342155 

1003 

657845 

36 

25 

332478 

956 

989721 

47 

342757 

1002 

657243 

35 

26 

333051 

954 

989693 

47 

343358 

1000 

656642 

34 

27 

333624 

953 

989665 

47 

343958 

999 

656042 

33 

28 

334195 

952 

989637 

47 

344558 

998 

655442 

32 

29 

334706 

950 

989609 

47 

345157 

997 

654843 

31 

30 
31 

335337 

949 

989582 
9.989553 

47 
47 

345755 
9.346353 

996 
994 

654245 
10.653647 

30 
29 

9.335906 

948 

32 

336475 

946 

989525 

47 

346949 

993 

653051 

28 

33 

337043 

945 

989497 

47 

347545 

992 

652455 

27 

34 

337610 

944 

989469 

47 

.348141 

991 

651859 

26 

35 

338176 

943 

989441 

47 

348735 

990 

651265 

25 

36 

338742 

941 

989413 

47 

349329 

988 

650671 

24 

37 

339306 

940 

989384 

47 

349922 

987 

650078 

23 

38 

.  339871 

939 

989356 

47 

350514 

986 

649480 

22 

39 

340434 

937 

989328 

47 

351106 

985 

64P894 

21 

40 
41 

340996 

936 

989300 
9.989271 

47 
47 

351697 

983 

982 

648303 

20 
19 

9.341558 

935 

9.352287 

10.647713 

42 

342119 

934 

989243 

47 

352876 

981 

647124 

18 

43 

342679 

932 

989214 

47 

353465 

980 

646535 

17 

44 

343239 

931 

989186 

47 

354053 

979 

645947 

16 

45' 

343797 

930 

9891.57 

47 

354640 

977 

645360 

16 

46 

344355 

929 

989128 

48 

355227 

976 

644773 

14 

47 

344912 

927 

989100 

48 

355813 

975 

644187 

13 

48 

345469 

926 

989071 

48 

356398 

974 

643602 

12 

49 

346024 

925 

989042  48  | 

356982 

973 

643018 

11 

50 

346579 

924 

989014 

48 

357566 

971 

642434 

10 

51 

9.347134 

922 

9.988985 

48 

9.358149 

970 

10.641851 

9 

62 

347687 

921 

988956 

48 

358731 

969 

641269 

8 

53 

348240 

920 

988927 

48 

359313 

968 

640687 

7 

54 

348792 

919 

988898 

48 

359893 

967 

^   640107 

6 

55 

349343 

917 

988869 

48 

360474 

966 

639526 

5 

66 

349893 

916 

988840 

48 

361053 

965 

638947 

4 

57 

350443 

915 

988811 

49 

361632 

963 

638368 

3 

58 

350992 

914 

988782 

49 

362210 

962 

637790 

2 

69 

351540 

913 

9887531  49  1 

r62787 

961 

637213 

1 

60 

352088 

911 

9887241  49  ' 

363364 

960 

6.36636 

0 

Li 

Cosine   j 

Sine   1   1 

CuUMg.     1 

1 

Tang   1  M.  | 

7T  Ufigrces. 


SINES  AND  TANGENTS.     ^^13  Degrees.) 


31 


I  Cosine   I  I).  I   T.iMg.   |  I).     \      Ci 


I  9.3-V20*o!8 
'   3.3263.= 


.373933 
374452 
374970 
375487 
376003 
376519 
377035 
377549 
378003 
378577 


.379089 
379601 
380 113 
380624 
381134 
381643 
382152 
3S2661 
383 16S 
383675 


898 
897 
896 
895 
893 
092 
891 
890 
889 
888 
887 
885 
884 
8S3 
882 
881 
880 
879 
87r 
876 
875 
874 
873 
872 
871 
870 
869 
867 
866 
865 


864 
883 
862 
861 
860 
859 
858 
857 
856 
854 


853 
852 
851 
850 
849 
848 
847 
846 
845 
844 


9.i 


.988103 
988073 
988043 
988013 
987983 
987953 
987922 
987892 
987862 
987832 


.987496 
987465 
987434 
987403 
987372 
987341 
997310 
987279 
987248 
987217 


49 

9.363304 

960 

49 

363940 

959 

49 

3645 1& 

958 

49 

365090 

957 

49 

365664 

955 

49 

368237 

954 

49 

366810 

953 

49 

367382 

952 

49 

367953 

951 

49 

368524 

950 

49 
49 

369094 

949 

9.369663 

948 

49 

370232 

946 

49 

370799 

945 

50 

371367 

944 

50 

371933 

943 

50 

372499 

942 

50 

373064 

941 

50 

373629 

940 

50 

374193 

939 

50 
50 

374756 

938 

9.375319 

937 

50 

375881 

935 

50 

376442 

934 

50 

377003 

933 

50 

377563 

932 

50 

378122 

931 

50 

378681 

930 

50 

379239 

929 

50 

379797 

928 

51 

380354 

927 

51 

9.380910 

926 

51 

381466 

925 

51 

382020 

924 

51 

382575 

923 

51 

383129 

922 

51 

383682 

921 

51 

384234 

920 

51 

384786 

919 

51 

385337 

918 

51 
51 

385888 

91V 

9.386438 

915 

51 

386987 

914 

51 

387536 

913 

52 

388084 

912 

52 

388631 

911 

52 

389178 

910 

52 

389724 

909 

52 

390270 

908 

52 

390815 

907 

52 
52 

391360 

906 

9.391903 

905 

52 

392447 

904 

52 

392989 

903 

52 

393531 

902 

62 

394073 

901 

52 

394614 

900 

52 

395154 

899 

52 

395694 

898 

{,2 

396233 

897 

52 

396771 

896 

10.6386361  »J0 
6360801  ?Q 
635485' :.<< 
6349101  57 
6343361 50 
6b3763'  55 
633190  54 


632618 
632047 
631476 

630908 

10^30337 
629768 
62920 1 
628633 
623067 
627501 
626936 
626371 
625807 
625244 


53 
52 
51 
50 
49 
48 
47 
46 
45 
44 
43 
42 
41 
40 
3& 
38 
37 
36 
35 
34 
33 
32 
31 
30 
29 
28 
27 
26 
25 
24 
23 
22 
21 
20 
107613562  19 

613013  18 

612464 

611918 

611369 

610822 

610276 

609730 

609185 

608640 


10.624881 
624119 
623558 
622997 
622437 
621878 
621319 
620781 
620203 
619848 


1.0.619090 
618534 
617980 
617425 
616871 
616318 
615768 
615214 
614683 
614112 


ov/ 


10.608097 
607553 
607011 
606489 
605927 
605386 
604846 
604306 
6037671 
6033291 


I      C  oine       I 


I  I 


r.-iiip.    I  M. 


EE* 


76  Degrees. 


sj 

(14  Degrees.;  a 

TABLE  OF  LOGARITHailC 

M. 

Sine 

1). 

Cosine   |  1). 

1   Tan-. 

D. 

Cotanj!.   i   1 

U 

9.383675 

844 

9. 9^6904 

52 

9.396771 

896 

10.6032^9 

60 

1 

384182 

843 

986873 

53 

397309 

896 

602691 

59 

2 

384687 

842 

986841 

53 

397846 

895 

602154 

58 

3 

385192 

841 

986809 

53 

398.383 

894 

601617 

5~ 

4 

385697 

840 

986778 

53 

398919 

893 

601081 

5d 

5 

386201 

^39 

986746 

53 

399455 

892 

600545 

55 

6 

386704 

838 

986714 

53 

399990 

891 

600010 

54 

7 

387207 

837 

986683 

53 

400524 

890 

599476 

53 

8 

387709 

836 

986651 

53 

401058 

889 

598942 

52 

9 

388210 

835 

986619 

53 

401591 

888 

598409 

51 

10 
11 

388711 

834 

986587 

53 
53 

402124 

887 
886 

597876 
10..597'344 

50 
49 

9.38^11 

833 

9.986555 

9.402656 

12 

389711 

832 

986523 

53 

403187 

885 

.596813 

48 

13 

390210 

831 

986491 

53 

403718 

884 

696282 

47 

14 

390708 

830 

986459 

53 

404249 

883 

59.5751 

46 

15 

391206 

828 

986427 

53 

404778 

882 

,595222 

45 

16 

391703 

827 

986395 

53 

405308 

881 

594692 

44 

17 

392199 

826 

986363 

54 

405836 

880 

594164 

43 

18 

392695 

825 

986331 

54 

406364 

879 

593636 

42 

19 

393191 

824 

986299 

54 

406892 

878 

593108 

41 

20 
21 

393685 
9.394179 

823 

822 

986266 
9.986234 

54 
54 

407419 

877 

592581 

40 
39 

9.407945 

876 

10.592055 

22 

394673 

821 

986202 

54 

408471 

875 

591529 

38 

23 

395166 

820 

986169 

54 

408997 

874 

591003 

37 

21 

395658 

819 

986137 

54 

409521 

874 

590479 

36 

25 

396150 

818 

986104 

54 

410045 

873 

589955 

35 

26 

396641 

817 

986072 

54 

410569 

872 

589431 

34 

27 

397132 

817 

986039 

54 

411092 

871 

588908 

33 

23 

397621 

816 

986007 

54 

411615 

870 

58.8385 

32 

29 

398111 

815 

935974 

54 

412137 

869 

587863 

31 

30 
3i 

398600 

814 

985942 

54 
55 

4126.58 

868 

587342 
10.. 586821 

30 

29 

9.399088 

813 

9.985909 

9.413179 

867 

32 

399575 

812 

985876 

55 

41.3699 

866 

586301 

28 

33 

400062 

811 

985843 

55 

414219 

865 

58578  I 

27 

34 

400549 

810 

985811 

55 

414738 

864 

585202 

26 

35 

401035 

809 

985778 

55 

41.5257 

864 

584743 

25 

3G 

401520 

808 

985745 

55 

415775 

863 

584225 

24 

37 

402005 

807 

985712 

55 

416293 

862 

583707 

23 

38 

402489 

806 

985679 

55 

416810 

861 

683190 

22 

33 

402972 

805 

985646 

55 

417326 

860 

582674 

21 

40 
41 

403455 

804 

985613 
9.985580 

55 
55 

417842 
9.418358 

859 

.582158 

20 
19 

9.403938 

803 

858 

10.581642 

42 

404420 

802 

985547 

55 

418873 

857 

581127 

18 

43 

404901 

801 

985514 

55 

419387 

S5G 

580613 

17 

44 

'  405382 

800 

985480 

55 

419901 

855 

580099 

16 

45 

405862 

799 

985447 

55 

420415 

855 

579.585 

15 

48 

406341 

798 

985414 

56 

420927 

854 

579073 

14 

47 

406820 

797 

985380 

56 

421440 

853 

578560 

13 

48 

407299 

796 

985347 

56 

421952 

852 

678048 

12 

49 

407777 

795 

985314 

56 

422463 

851 

577537 

11 

50 
51 

408254 

794 

985280 
9.985247 

56 
56 

422974 

850 

577026 
10.576516 

10 
9 

9.408731 

794 

9.423484 

849 

52 

409207 

793 

985213 

56 

423993 

848 

576007 

8 

53 

409682 

792 

985180 

56 

424503 

848 

575497 

7 

54 

410157 

7!)1 

985146 

56 

42.5011 

847 

574989 

6 

55 

410632 

790 

985113 

56 

425519 

846 

674481 

5 

5fi 

411106 

789 

985079 

56 

426027 

845 

573973 

4 

57 

411579 

788 

985045 

56 

4265.34 

844 

573466 

3 

58 

412055i 

787 

985011 

56 

427041 

843 

572959 

2 

59 

412524 

786 

984978 

58 

427547 

843 

572453 

1 

fiO 

412996 

785 

994944 

58 

428052 

842 

.57194^ 

0 

1 

Oiisiiie 

riinn   | 

1  C.tu,.  1 

1   Tang   1  M.  | 

75  Degrees. 


SINES  AND  TANGENTS.   (15  Degrees.) 

33 

M. 

1    Sine 

1   0. 

1  Cosine   1  D. 

1   Tan!.'. 

1   1>. 

!   Cotaiii:.   j 

"U 

9.412996 

785 

9.984944 

'  .57 

9.428052 

842 

10.571948  60 

1 

413467 

784 

984910 

57 

428557 

841 

571443  b^ 

2 

413938 

783 

984876 

57 

429062 

840 

570938  !'^ 

3 

414408 

783 

984842 

57 

429666 

839 

570434  ■  57 

4 

414878 

782 

984808 

57 

430070 

838 

569930  :  56 

5 

415347 

781 

984774 

57 

430573 

838 

569427 

55 

6 

415815 

780 

984740 

57 

431075 

837 

568925 

54 

7 

416283 

779 

9S4706 

57 

431577 

836 

568423 

53 

8 

416751 

778 

984672! 57 

432079 

835 

667921 

62 

9 

417217 

777 

984637  57 

432580 

834 

567420 

51 

10 

417684 

770 

9846031 57 

4330S0 

833 

566920 

50 

11 

9.418150 

775 

9.984569! 67 

'9.433580 

832 

10.566420 

49 

12 

418615 

774 

984535 

57 

434080 

832 

565920 

48 

13 

419079 

773 

984500 

57 

434579 

831 

66.5421 

47 

14 

419544 

773 

984466 

57 

435078 

830 

564922 

46 

15 

420007 

772 

984432 

58 

435676 

829 

564424 

45 

16 

420470 

771 

984397 

58 

436073 

828 

563927 

44 

17 

420933 

770 

984363 

58 

436670 

828 

563430 

43 

18 

421395 

769 

984328 

58 

437067 

827 

662933 

42 

19 

421857 

768 

984294 

58 

437563 

826 

562437 

41 

20 

422318 

767 

984259 

58 

438059 

'82.'> 

561941 

40 

21 

9  422778 

767 

9.984224 

58 

9.438554 

824 

I0r561446 

.39 

22 

423238 

766 

984190 

58 

439048 

823 

560952 

38 

23 

423697 

765 

984155 

58 

439543 

823 

560457 

37 

24 

424156 

764 

984120 

58 

440036 

822 

569964 

36 

25 

424615 

763 

984085 

58 

440529 

821 

5.59471 

35 

26 

425073 

762 

984050 

68 

441C22 

820 

658978 

34 

27 

425530 

761 

984015 

58 

441514 

819 

558486 

33 

28 

425987 

760 

983981  58 

442006 

819 

557994 

32 

29 

426443 

760 

983946  58 

442497 

818 

657503 

31 

30 
31 

426899 

759 

083911  .58 
9.983875158 

442988 

817 

5.57012 

30 

29 

9.427354 

758 

9.'W3479 

816 

10.556521 

32 

427809 

757 

983840  59 

443968 

816 

556032 

28 

33 

428263 

756 

983805  59 
983770  59 

444458 

815 

555542 

27 

34 

428717 

755 

444947 

814 

565053 

26 

35 

429170 

754 

9»3735  59 

445435 

813 

654565 

25 

36 

429623 

753 

9837001  59 

445923 

812 

654077 

24 

37 

430075 

752 

983664 

59 

446411 

812 

553n89 

23 

38 

430527 

752 

983629 

59 

446898 

811 

553102 

22 

39 

430978 

751 

983594 

59 

447384 

810 

552616 

21 

40 

4f 

431429 

750 

983558 

59 
59 

447870 

809 

552130 

20 
19 

9.431879 

749 

9.983523 

9.448.356 

809 

10.. 55 1644 

42 

432329 

749 

98.3487 

59 

448841 

808 

6511.59 

18 

43 

432778 

748 

983452 

59 

449326 

807 

660674 

17 

44 

433226 

747 

9834161  59  I 

449810 

806 

550190 

16 

45 

433675 

746 

983381 

59 

450294 

806 

.549706 

16 

46 

434122 

745 

983345 

59 

460777 

805 

649223 

14 

47 

434569 

744 

983309 

59 

451260 

804 

.548740 

13 

48 

435016 

744 

9832731  60 

451743 

803 

548257 

•12 

49 

435462 

743 

983238  60 

462225 

802 

547775 

11 

50 
51 

435908 
9.436353 

742 

983202 
9.983166 

60 
60 

452706 

802 

647294 
10.546813 

10 
9 

741 

9.4.53187 

801 

52 

436798 

740 

983130 

60 

453668 

SOO 

546332 

8 

53 

437242 

740 

983094 

60 

464148 

799 

645852 

7 

54 

437686 

739 

983058 

60 

454628 

799 

545372 

6 

55 

438129 

738 

983022 

60 

455107 

798 

641893 

6 

56 

438572 

737 

982986  60 

455586 

797 

614414 

4 

57 

439014 

736 

982950  60 

466064 

•  796 

543936 

3 

58 

439456 

736 

982914  60 

456542 

796 

543458 

2 

59 

439897 

735 

9828781 60 

4.57019 

795 

542981 

60 

440338 

734 

9828421 60 

457496 

794 

542504 

0 



Cosine   | 

1 

Sine   1   1 

Colling.  1 

Taxa.   1  M.  j 

74  Degrees. 


34 

(16  Degrees.)  a 

TAHLE  OF  LOGARITHMIC 

M. 

Sine 

D. 

VoifUic 

I). 

Tang. 

D. 

Cot:ir</   j   1 

^ 

9.440338 

734 

9. 98:..' 842 

60 

9.457496 

794 

"107542504 

lo 

1 

440778 

733 

982805 

60 

457973 

793 

542027 

59 

2 

441218 

732 

982769 

61 

458449 

793 

541551 

58 

3 

441658 

731 

982733 

61 

458925 

792 

541075 

57 

4 

442090 

731 

982696 

61 

459400 

791 

540600 

56 

5 

442535 

730 

982660 

61 

459875 

790 

540125 

55 

6 

442973 

729 

982624 

61 

460349 

790 

639651 

54 

7 

443410 

728 

982587 

61 

460323 

789 

539177 

53 

8 

443847 

727 

982551 

61 

461297 

788 

538703 

52 

9 

444284 

727 

982514 

61 

461770 

788 

533230 

51 

10 
11 

444720 

726 

982477 

61 
61 

462242 

787 

537758 

50 
49 

9.445155 

725 

9.982441 

9.462714 

786 

10.537286 

12 

445590 

724 

982404 

61 

463186 

785 

530814 

48 

I'd 

446025 

723 

9823G7  61 

463658 

785 

536342 

47 

14 

446459 

723 

982301 

61 

464129 

784 

.535871 

46 

15 

446893 

722 

982294 

61 

464599 

783 

5J5401 

45 

16 

447326 

721 

982257 

61 

46506: 

783 

rm^jn 

44 

17 

447759 

720 

982220 

62 

465539 

782 

534461 

43 

18 

448191 

720 

982183 

62 

466008 

-781 

533992 

42 

19 

448623 

719 

982146 

62 

466476 

780 

533524 

41 

20 

21 

449054 
9.449485 

718 
717 

982109 
9.982072 

62 
62 

466945 

780 

533055 
10.532587 

40 
39 

9.467413 

779 

22 

449915 

716 

982035 

62 

467880 

778 

532120 

38 

23 

450345 

716 

981998 

62 

468347 

778 

53?653 

37 

24 

450775 

715 

981961 

62 

468814 

777 

531186 

36 

25 

451204 

714 

981924 

62 

469280 

776 

530720 

35 

26 

451632 

713 

981886 

62 

469746 

775 

530254 

34 

27 

452060 

713 

981849 

62 

470211 

775 

529789 

33 

28 

452488 

712 

981812 

62 

470676 

774 

529324 

32 

29 

452915 

711 

981774 

62 

471141 

773 

528859 

31 

30 

453342 

710 

981737 

62 

471605 

773 

5283iJ5 

30 

31 

9.453768 

710 

9.981699 

63 

9.472008 

772 

10.527932 

¥j 

32 

454194 

709 

981662 

63 

472532 

771 

527468 

28 

33 

454619 

708 

981625 

63 

472995 

771 

527005 

27 

34 

455044 

707 

981587 

63 

473457 

770 

526543 

26 

35 

455469 

707 

981549 

63 

473919 

769 

.526081 

25 

36 

455893 

706 

981512 

63 

474381 

769 

52.5619 

24 

37 

456316 

705 

981474 

63 

474842 

768 

5251.58 

23 

38 

456739 

704 

981436 

63 

475303 

767 

524697 

22 

39 

457162 

704 

981399 

63 

475763 

767 

524237 

21 

40 

457584 

703 

981361 

63 

476223 

766 

523777 

20 

41 

9.458006 

702 

9.981323 

63 

9.476683 

765 

10.523317 

19 

42 

458427 

701 

981285 

63 

477142 

765 

522858 

18 

43 

458848 

70.1 

981247 

63 

477601 

764 

522399 

17 

44 

459268 

700 

981209 

63 

478059 

763 

521941 

16 

45 

459688 

699 

981171 

63 

478517 

763 

521483 

15 

46 

460108 

698 

981133 

64 

478975 

762 

.521025 

14 

47 

460527 

698 

981095 

64 

479432 

761 

52056S 

13 

48 

460946 

697 

981057 

64 

479889 

761 

520111 

12 

49 

461364 

696 

981019 

64 

480345 

760 

519655 

11 

50 
51 

461782 

695 

980Q81 
9.980942 

64 
64 

480801 

759 

519199 
10.518743 

10 
9 

9.462199 

695 

9.481257 

759 

52 

462616 

694 

980904 

64 

481712 

758 

518288 

8 

53 

463032 

693 

980866 

64 

482167 

757 

^  517833 

7 

54 

463448 

693 

980^27 

64 

482621 

757 

617379 

6 

55 

46386^. 

692 

980789 

64 

483075 

756 

516925 

5 

56 

464279 

691 

980750!  64 

483529 

755 

616471 

4 

57 

464694 

690 

•  9807 12 

64 

483982 

755 

616018 

3 

58 

465108 

69a 

980673 

64 

484435 

754 

615565 

2 

59 

465522 

68» 

980635 

64 

484887 

7i=a 

615113 

1 

60 

465935 

688 

98059b i  64 

485339 

753 

5T4f>fi1 

0 

^ 

Coi-iiie 

1   Sine   1 

Coluiig. 

1    Tang.   |M.| 

73-  l)<igitesk 


SINES  AND  TANGENTS. 

(l7  Degrees 

0 

35 

M^ 

Si  IK!   1 

n    ! 

Cosine   |  !). 

Ttuig. 

U. 

Ootan^.  .  1   1 

0 

9.46G935 

688 

9.980596 

64 

9.485339 

755 

10.514661  i  60  1 

1 

466348 

688 

980558 

64 

485791 

752 

614209 

C9 

2 

466761 

087 

980519 

65 

486242 

751 

5137i38 

58 

3 

467173 

686 

980480 

65 

486693 

751 

613307 

57 

4 

467585 

685 

980442 

65 

487143 

750 

5128.57 

56 

5 

467996 

685 

980403 

65 

487593 

749 

612407 

55 

6 

468407 

684 

980364 

65 

488043 

749 

611957 

54 

7 

468817 

683 

980325 

65 

488492 

748 

511.508 

53 

8 

469227 

683 

980286 

65 

488941 

747 

6110.59 

52 

9 

469637 

682 

980247 

65 

489390 

747 

510610 

51 

10 

470046 

681 

980208 

65 

489838 

746 

610162 

50 

11 

9.470455 

6^0 

9.980169 

65 

9.490286 

746 

10  .009714 

49 

12 

470863 

680 

980130 

65 

490733 

745 

509267 

48 

13 

471271 

679 

980091 

65 

491180 

744 

608820 

47 

14 

471^9 

678 

980052 

65 

491627 

744 

608373 

46 

15 

472086 

678 

980012 

65 

492073 

743 

607927 

45 

16 

472492 

677 

979973 

65 

492519 

743 

507481 

44 

17 

47289S 

676 

979934 

66 

492965 

742 

6070.35 

43 

18 

473304 

676 

979895 

66 

49.3410 

741 

606590 

42 

19 

473710 

675 

979855 

66 

493854 

740 

606148 

41 

20 

474115 

674 

979816 

66 

494299 

740 

505701 

40 

21 

9.474519 

674 

9.979776 

66 

9.494743 

740 

10.50.5257 

39 

22 

474923 

673 

979737 

66 

496186 

739 

604814 

88 

23 

475327 

672 

979697 

66 

495630 

738 

504370,371 

24 

475730 

672 

979668 

66 

496073 

737 

503927 

36 

25 

476133 

671 

979618 

66 

496615 

737 

503485 

35 

26 

476536 

670 

979.'i79 

66 

496957 

736 

503043 

34 

:i7 

476938 

669 

979539 

66 

497399 

736 

602601 

33 

28 

477340 

669 

979499 

66 

497841 

735 

•  502159 

32 

29 

177741 

668 

979459 

66 

498282 

734 

501718 

31 

30 

478142 

667 

979420 

66 

498722 

734 

601278 

30 

31 

9.478542 

667 

9.979380 

66 

9.499163 

733 

10.500837 

29 

32 

478942 

666 

979340 

66 

499603 

733 

600397 

28 

33 

479342 

665 

979300 

67 

500042 

732 

499958 

27 

34 

479741 

665 

979260 

67 

500481 

731 

499519 

26 

35 

480140 

664 

979220 

67 

500920 

731 

499080 

25 

36 

480539 

6  53 

979180 

67 

601359 

730 

498641 

24 

37 

480937 

663 

979140 

67 

601797 

730 

498203 

23 

38 

481334 

662 

979100 

67 

602235 

729 

497765  !  22  | 

39 

481731 

661 

979059 

67 

602672 

728 

497328  j  21  ! 

40 

482128 

66i 

979019 

67 

603109 

728 

496891  i20| 

41 

9.482525 

600 

9.978979 

67 

9.603546 

727 

10.496454 

10 

42 

482921 

659 

978939 

67 

503982 

727 

496018 

18 

43 

483316 

659 

978898 

67 

604418 

726 

495582 

17 

44 

483712 

658 

978858 

67 

504854 

725 

495146 

16 

45 

484107 

657 

978817 

67 

506289 

725 

494711 

15 

46 

484501 

657 

978777 

67 

605724 

724 

494276 

14 

47 

484895 

656 

978736 

67 

.6061.59 

724 

493841 

13 

48 

485289 

656 

978696 

68 

506593 

723 

493407 

12 

49 

485682 

655 

978655 

68 

507027 

722 

492973 

.11 

50 

486075 

654 

978615 

68 

607460 

722 

492.540 

10 

51 

9.486467 

653 

9.978.574 

68 

9.507893 

721 

10.492107 

9 

52 

486860 

653 

978.533 

68 

508326 

721 

491674 

8 

53 

487251 

652 

978493 

68 

508759 

720 

491241 

7 

54 

487643 

651 

978452 

68 

509191 

719 

490809 

6 

55 

488034 

651 

978411 

68 

509622 

719 

490378 

5 

56 

488424 

050 

978370 

68 

5100.54 

718 

489940 

4 

57 

488814 

650 

978329 

68 

510485 

"18 

489515   3 

58 

489204 

649 

978288 

68 

610916 

717 

489084   2 

59 

489593 

648 

978247 

68 

511346 

710 

48R6,54   1 

60 

489982 

648 

978206 

68 

611776 

716 

488224'  0 

~ 

Ciit-iiie 

Sine   j 

Coiani'. 

1    ■!  ann   j  M. 

18 


71  Uegiees. 


36 

ri  8  Degrees.)  a 

TAWiB  OF  LOGARITHMIC 

"m" 

1   Sine 

1   I>. 

1  Cosine   |  1). 

1   'I'aiig. 

1   D 

1   Cutaiiu. 

0 

9.489982 

648 

9.978206168 

9.511776 
512206 

1  716 

10.488224  1  60 

1 

490371 

648 

978165 

68 

716 

487794 

59 

2 

490759 

647 

978124 

68 

512635 

715 

487365 

58 

3 

491147 

646 

978083 

69 

513064 

714 

486936 

57 

4 

491535 

646 

978042 

69 

513493 

714 

486507 

56 

5 

491922 

645 

978001 

69 

613921 

713 

486079 

55 

6 

492308 

644 

977959 

69 

614349 

713 

485651 

54 

7 

492695 

644 

977918 

69 

614777 

712 

485223 

53 

8 

493081 

643 

977877 

69 

615204 

712 

484796 

52 

9 

493466 

642 

977835 

69 

615631 

711 

484369 

51 

10 
11 

493851 

642 

977794 
9.977752 

69 
69 

516057 
9.616484 

710 

483943 
10.483516 

50 
49 

9.494236 

641 

710 

12 

494621 

641 

9V7711 

69 

516910 

709 

483090 

48 

13 

495005 

640 

977669 

69 

517335 

709 

482665 

47 

14 

495388 

639 

977628 

69 

517761 

708 

482239 

46 

15 

495772 

639 

977586 

69 

618185 

708 

481815 

45 

16 

496154 

638 

977544 

70 

618610 

707 

481390 

44 

17 

496537 

637 

977503 

70 

519034 

706 

480966 

43 

18 

496919 

637 

977461 

70 

519458 

706 

480542 

42 

19 

497301 

636 

977419 

70 

519882 

705 

480118 

41 

20 
21 

497682 

636 

977377 

70 
70 

620305 

705 

479695 
10.479272 

40 
39 

9.498064 

635 

9.977.335 

9.520728 

704 

22 

498444 

634 

977293 

70 

621151 

703 

478849 

38 

23 

498825 

634 

977251 

70 

621573 

703 

478427 

37 

24 

499204 

633 

977209 

70 

621995 

703 

478005 

36 

25 

499584 

632 

977167 

70 

622417 

702 

477583 

35 

26 

499963 

632 

977125 

70 

522838 

702 

477162 

34 

27 

500342 

631 

977083 

70 

523259 

701 

476741 

33 

28 

600721 

631 

977041 

70 

523680 

701 

476320 

32 

29 

501099 

630 

976999 

70 

524100 

700 

475900 

31 

30 
31 

501476 

629 

976957 

70 
70 

524520 

699 

475480 

30 
29 

9.501854 

629 

9.976914 

9.524939 

699 

10.475061 

32 

502231 

628 

976872 

71 

525359 

698 

474641 

28 

33 

502607 

628 

9768.30 

71 

525778 

698 

474222 

27 

34 

502984 

627 

976787 

71 

526197 

697 

473803 

26 

35 

503360 

026 

976745 

71 

526615 

697 

473385 

25 

36 

503735 

626 

976702 

71 

527033 

696 

472967 

24 

37 

504110 

625 

976660 

71 

527451 

696 

472549 

23 

38 

504485 

625 

976617 

71 

527868 

695 

472132 

22 

39 

504860 

624 

976574 

71 

628285 

695 

471715 

21 

40 

41 

505234 

623 

976532 
9.976489 

71 
71 

628702 

694 

4712118 
0.470881 

20 
19 

9.505608 

623 

9.529119 

693 

42 

505981 

622 

976446 

71 

529535 

693 

470465 

18 

43 

606354 

622 

976404 

71 

629950 

693 

470050 

17 

44 

506727 

621 

976361 

71 

630366 

692 

469634 

16 

45 

507099 

620 

976318 

71 

530781 

691 

469219 

15 

46 

507471 

620 

976275 

71 

531196 

691 

468804 

14 

47 

507843 

619 

976232 

72 

631611 

690 

468389 

13 

48 

508214 

619 

976189 

72 

532025 

690 

467975 

12 

49 

508585 

618 

976146 

72 

532439 

689 

467561 

11 

50 
51 

508956 

618 

976103 
9.976060 

72 
72 

532853 

689 

467147 

10 
9 

9.509326 

617 

9.533266 

688 

10.466734 

52 

509696 

6]6 

976017 

72 

533679 

688 

466321 

8 

53 

510065 

616 

975974 

72 

534092 

687 

465908 

7 

54 

510434 

615 

975930 

72 

534504 

687 

,  465496 

8 

55 

510803 

615 

975887 

72 

634916 

686 

465084 

5 

56 

511172 

614 

975844 

72 

535328 

686 

464672 

4 

57 

511540 

613 

975800 

72 

535739 

685 

464361 

3 

C'8 

611907 

613 

975757 

72 

536150 

685 

463850 

2 

59 

612275 

612 

97.5714 

72 

536561 

684 

403439 

1 

60 

512642 

G12 

975670 

72 

536972 

684 

463028 

0 

J. 

Cosine 

1 

Sine   1 

(J(Kan<r. 

'J'aiifr.    1  M.  1 

71  De;!j;ree8. 


SINES  AND  TANGENTS.   (19  Degrees.) 

S7 

M. 

Sine 

D. 

Cosine  1  D. 

'I'itng. 

n. 

Coiatif.   ! 

"IT 

9.512642 

612 

9.975670 

73 

9.536972 

684 

10.463028  60 

1 

513009 

611 

975627 

73 

537382 

683 

462618  .59 

2 

513375 

611 

975583 

73 

537792 

683 

462208 

68 

3 

513741 

610 

975539 

73 

538202 

682 

461798 

67 

4 

514107 

609 

975496 

73 

538611 

682 

461389 

56 

5 

514472 

609 

976452 

73 

639020 

681 

460980 

55 

6 

514837 

608 

9/5408 

73 

639429 

681 

460671 

54 

7 

515202 

608 

975366 

73 

639837 

680 

460163 

53 

8 

515566 

607 

976321 

73 

.540246 

680 

459756 

62 

9 

615930 

607 

976277 

73 

640653 

679 

469347 

51 

}0 
'll 

516294 

606 
605 

975233 

73 
73 

.541061 

679 

4.58939 
10.4.586.32 

60 
49 

9.516657 

9.975189 

9.641468 

678 

12 

517020 

605 

975145 

73 

641875 

678 

458125 

48 

13 

517382 

604 

975101 

73 

.542281 

677 

457719 

47 

14 

517745 

604 

975057 

73 

542688 

677 

457312 

46 

15 

518107 

603 

97.5013 

73 

643094 

676 

456906 

45 

16 

518468 

603 

974969 

74 

643499 

676 

456601 

44 

17 

518829 

602 

974925 

74 

543905 

675 

456095 

43 

18 

519190 

601 

974880 

74 

544310 

675 

455690 

42 

19 

519551 

601 

974836 

74 

544715 

674 

■  456285 

41 

20 
21 

519911 

600 

974792 

74 

74 

546119 

674 
673 

4,54881 
10.4.544V6 

40 
39 

9.520271 

600 

9.974748 

9.. 546524 

22 

520631 

599 

974703 

74 

545928 

673 

454072 

38 

23 

520990 

599 

974659 

74 

546331 

672 

463669 

37 

24 

521349 

598. 

974614 

74 

546736 

672 

453265 

36 

25 

.521707 

598 

974570 

74 

647138 

671 

452862 

35 

26 

522066 

597 

974525 

74 

647540 

671 

452460 

34 

27 

522424 

596 

974481 

74 

647943 

670 

452067 

33 

28 

522781 

596 

974436 

74 

648346 

670 

461655 

32 

29 

.523138 

595 

974391 

74 

548747 

669 

451253 

31 

30 

523495 

595 

974347 

75 

649149 

669 

4.50861 

30 

31 

9.. 523852 

594 

9.974302 

75 

9.549550 

668 

10.450450 

29 

32 

524208 

594 

974267 

76 

649951 

668 

460049 

28 

33 

524564 

593 

974212 

76 

550352 

667 

449648 

27 

34 

524920 

593 

974167 

76 

5507.52 

667 

449248 

26 

35 

525275 

592 

974122 

75 

651152 

666 

448848 

25 

36 

525630 

591 

974077 

75 

651552 

666 

448448 

24 

37 

525984 

591 

974032 

75 

551952 

666 

448048 

23 

38 

526339 

590 

973987 

75 

562351 

665 

447649 

22 

39 

526693 

590 

973942 

75 

652750 

666 

447250 

21 

40 
41 

527046 

589 

973897 
9.973852 

75 
75 

553149 
9.5.53548 

664 
664 

446851 

20 
19 

9.527400 

589 

10.446452 

42 

527753 

588 

973807 

75 

563946 

663 

446054 

18 

43 

.528105 

588 

973761 

76 

554344 

663 

445656 

17 

44 

528458 

587 

973716 

76 

554741 

662 

445259 

16 

45 

.528810 

587 

973671 

76 

655139 

662 

444861 

16 

46 

529161 

586 

973625 

76 

556536 

661 

444464 

14 

47 

629513 

586 

973580 

76 

555933 

661 

444067 

13 

48 

529864 

585 

973535 

76 

556329 

660 

443671 

12 

49 

.^^302 15 

585 

973489 

76 

556725 

660 

443275 

11 

50 
51 

b30565 
1/530915 

584 

973444 

76 
76 

557121 

659 

442879 
10.442483 

10 
9 

584 

9.973398 

9.667517 

659 

62 

531265 

583 

973352 

76 

557913 

659 

442087 

8 

53 

531614 

582 

973307 

76 

658308 

658 

441692 

7 

54 

531963 

682 

973261 

76 

568702 

658 

441298 

6 

55 

532312 

581 

973215 

76 

559097 

657 

440903 

5 

56 

532661 

581 

973169 

76 

569491 

667 

440509 

4 

57 

633009 

580 

973124 

76 

659885 

656 

440115 

3 

58 

533357 

680 

973078 

76 

560279 

656 

439721 

3 

59 

533704 

679 

973032  77 

500673 

655 

439327 

1 

il 

534052 

578 

972986  77 

561066 

655 

438934 

0 

^ 

Cosine 

1 

Sine    1 

Cotang. 

1 

Tang.   1  M.  | 

70  Degrees 


38 

\^20  Degrees.}  a 

TABLE  OF  LOGARITHMIC 

1   Sine 

1   D. 

1  <;osine   |  U 

TaiiR. 

}  » 

1  C.iju.a.   1   1 

~0 

9.534052 

1  578 

9.9729S6 

l77 

"9.561066 

655 

10.43.Si^34 

60 

1 

534399 

577 

972940 

77 

561459 

6.64 

438541 

59 

2 

534745 

577 

972894 

77 

561851 

654 

438149 

68 

3 

535092 

577 

972848 

77 

562244 

663 

437756 

o7 

4 

535438 

676 

972802 

77 

562636 

663 

437364 

56 

s 

535783 

676 

972766 

77 

663028 

653 

436972 

55 

6 

536129 

576 

972709 

77 

66.3419 

652 

436681 

54 

7 

536474 

574 

972663 

77 

563811 

652 

436189 

53 

8 

536818 

574 

972617 

77 

564202 

661 

435798 

52 

9 

537163 

673 

972570 

77 

564592 

661 

436408 

51 

10 

537507 

573 

972624 

77 

564983 

660 

435017 

50 

11 

9.537851 

572 

9:972478 

77 

9.. 5653713 

650 

10.434627 

49 

12 

538194 

672 

972431 

78 

565763 

649 

434237 

48 

13 

538538 

671 

972386 

78 

666153 

649 

433847 

47 

14 

538880 

671 

972338 

78 

566542 

649 

433468 

46 

15 

539223 

670 

972291 

78 

566932 

648 

433068 

45 

16 

539565 

670 

972246 

78 

567320 

648 

432680 

44 

17 

539907 

669 

972198 

78 

667709 

647 

432291 

43 

18 

540249 

569 

972161 

78 

668098 

647 

431902 

42 

19 

540590 

568 

972105 

79 

668486 

646 

431514 

41 

20 

21 

540931 
9.541272 

568 

972068 
9.972011 

78 
78 

568873 
9.669261 

646 

431127 
10.430739 

40 
39 

567 

645 

22 

541613 

567 

971964 

78 

669648 

645 

430352 

38 

23 

541953 

666 

971917 

78 

570035 

645 

429965 

37 

24 

542293 

566 

971870 

78 

570422 

644 

429578 

36 

25 

542632 

505 

971823 

78 

570809 

644 

429191 

35 

26 

542971 

565 

971776 

78 

571195 

643 

428805 

34 

27 

543310 

564 

971729 

79 

571681 

643 

428419 

33 

28 

.543649 

564 

971682 

79 

671967 

642 

428033 

32 

29 

543987 

563 

971635 

79 

572352 

642 

427648 

31 

30 

541325 

.663 

971588 

79 

572738 

642 

427262 

30 

31 

9.544663 

662 

9.971540 

79 

9.573123 

641 

10.426877 

29 

32 

545000 

562 

971493 

79 

673507 

641 

426493 

28 

33 

545338 

561 

971446 

79 

573892 

640 

426108 

27 

34 

545674 

561 

971398 

79 

574276 

640 

425724 

26 

35 

646011 

560 

971351 

79 

674660 

639 

425340 

25 

36 

,546347 

560 

971303 

79 

675044 

639 

424950 

24 

37 

546683 

559 

971256 

79 

675427 

639 

42457^3 

23 

38 

547019 

669 

971208 

79 

575810 

638 

424190 

22 

39 

547354 

558 

971101 

79 

576193 

638 

423807 

21 

40 

41 

547689 

658 

971113 
9.971066 

79 

80 

576576 

637 

423424 

20 
19 

9.648024 

557 

9.576958 

637 

10.423041 

42 

648359 

557 

971018 

80 

577341 

G36 

422659 

18 

43 

548693 

656 

970970 

80 

677723 

636 

422277 

17 

44 

549027 

660 

970922 

80 

678104 

636 

421896 

16 

45 

649360 

555 

970874 

80 

678486 

635 

421614 

15 

46 

549693 

655 

970827 

80 

678867 

635 

421133 

14 

47 

550026 

654 

970779 

80 

679248 

634 

420752 

13 

48 

560359 

554 

970731 

80 

679629 

634 

420371 

12 

49 

660692 

553 

970683 

80 

580009 

634 

419991 

11 

50 

551024 

663 

970635 

80 

580389 

633 

419611 

10 

51 

9.551356 

552 

9.970586 

80 

9.580769 

633 

10.419231 

9 

52 

661687 

552 

970538 

80 

581149 

632 

418851 

8 

53 

652018 

5.62 

970490 

80 

681528 

632 

418472 

7 

54 

652349 

551 

970442 

80 

581907 

632 

,  418093 

6 

55 

662680 

561 

970394 

80 

582286 

631 

417714 

5 

56 

663010 

650 

970346 

81 

682666 

631 

417335 

4 

57 

653341 

550 

970297 

81 

683043 

630 

4169.57 

3 

58 

553670 

649 

970249 

81 

583422 

630 

416578 

2 

59 

554000 

549 

970209 

81 

683800 

629 

416200 

1 

60 

554329 

548 

970152 

81 

684177 

629 

41.5823 

0 

Ouhiuii 

1   «'''^   1   1 

Coiang. 

Tang.   1  M.  | 

Degrtes. 


SINES  AND  TANGENTS 

.  (21  Degrees 

•; 

39 

JJ: 

Sine   1 

».    ! 

Cosine  1  I).  1 

'J'ni.e.   1   I).   1 

CotaiiL'  1 

0 

y.5543x;y 

548  1 

9.970152  81 

9.584177 

629 

10.415823  1 

60 

654658 

648 

970103  81 

684555 

629 

415445 

59 

2 

554987 

547 

970055  81 

584932 

628 

41.5068 

58 

3 

555315 

547 

970006  81 

585309 

628 

414691 

57 

4 

655643 

546 

969957 

81 

685686 

627 

414314 

56 

5 

555971 

646 

969909 

81 

586062 

627 

413938 

55 

8 

556299 

646 

969860 

81 

686439 

627 

41.3.561 

64 

7 

556626 

645 

969811 

81 

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663707 

5.54 

336293 

15 

46 

'  622135 

456 

9580961  97 

664039 

553 

33.5961 

14 

47 

622409 

456 

9.58038 

97 

664371 

653 

335629 

13 

48 

622682 

455 

957979 

97 

664703 

553 

3.3.5297 

12 

49 

622956 

455 

957921 

97 

665035 

553 

3.34965 

11 

50 
51 

623229 

455 

957863 
9.957804 

97 
97 

665366 
9.66.5697 

5.52 
552 

334634 
1(573.34303 

10 
9 

9.623502 

454 

52 

623774 

454 

957746!  98 

666029 

552 

333971 

8 

53 

624047 

454 

957687 

98 

666360 

551 

333640 

7 

54 

624319 

453 

957628 

98 

666691 

651 

333309 

6 

55 

624591 

453 

957570 

98 

667021 

.551 

'  332979 

6 

56 

624863 

453 

957511 

98 

667352 

651 

332648 

4 

57 

62513.'i 

452 

957452 

98 

66768:^ 

6.50 

332318 

3 

58 

625406 

452 

957393 

98 

668013 

550 

331987 

2 

59 

625677 

452 

gi>7335 

98 

668.343 

550 

3316.57 

1 

60 

6259 Ifi 

451 

957276; 98 

668672 

550 

331?28 

0 

1   Cusiiie 

1   Si.e    1 

1   CMilllt! 

1 

1   'lixiii.  |.M.  1 

t5  Dtgr*^* 


-4/'' 


SINFS  AND  TA^'^1.^-TS 

(S5  Degrees.) 

43 

SI. 

Siiu! 

1). 

v;'..<iiie  1  1). 

1   'i'siiij;. 

n. 

1   I'orjiiisi.  j 

~ir 

'9T625948 

451 

9.95VV76 

98 

9.668673 

5.50 

10. 33 1327) 60 

1 

626219 

451 

957217 

98 

669002 

549 

330998  :" 

2 

626490 

451 

9571.58 

98 

669332 

549 

330668 

r.6 

3 

626760 

450 

957099 

98 

669661 

549 

330339 

57 

4 

627030 

450 

95";  040 

98 

669991 

548 

3*^0009 

56 

T) 

627300 

450 

95698 1 

98 

670320 

548 

329680 

55 

fi 

627570 

449 

956921 

99 

670649 

548 

329351 

64 

7 

627840 

449 

956862 

99 

670977 

548 

329023 

53 

8 

628109 

449 

956803 

99 

671.306 

547 

328894 

52 

9 

628378 

448 

936744 

99 

671634 

547 

328366 

51 

10 

628647 

448 

956684 

99 

671963 

547 

328037 

50 

11 

9.628916 

447 

9.950625 

99 

9.672291 

547 

10.327709 

49 

12 

629185 

447 

956566 

99 

672619 

646 

327.381 

48 

13 

629453 

447 

956.^^06 

99 

672947 

.546 

327053 

47 

R 

629721 

446 

9.56447 

99 

673274 

546 

326726 

46 

15 

6299S9 

446 

956387 

99 

673602 

540 

326398 

45 

16 

630257 

446 

956327 

99 

673929 

545 

326071 

44 

17 

630524 

446 

956268 

99 

074257 

545 

326743  43  1 

18 

630792 

445 

956208 

100 

6V4584 

645 

325416 

42 

19 

631059 

445 

956148 

100 

674910 

544 

325090 

41 

20 

21 

631326 
9.031593 

445 

956089 

100 
100 

675237 
9.675564 

544 
644 

324763 

40 
39 

444 

9.9.56029 

10.324436 

22 

631859 

444 

955969 

100 

675890 

644 

324110 

38 

23 

632125 

444 

955909 

100 

676216 

643 

323784 

37 

24 

. 032392 

443 

955849 

100 

676543 

.543 

323457 

36 

2o 

632658 

443 

955789 

100 

6768'G9 

643 

323131 

35 

26 

632923 

443 

955729 

100 

677194 

643 

322806 

34 

27 

633189 

442 

955f.6'J 

100 

677.'S20 

642 

322480 

33 

28 

633454 

443 

955609 

100 

677846 

642 

322154 

32 

29 

633719 

442 

955548 

100 

678171 

642 

321829 

31 

30 

633984 

441 

95.5488 

100 

678496 

642 

321504 

30 

31 

9.634249 

441 

9.95.5428 

101 

9.678821 

641 

10.321179 

29 

32 

634514 

440 

955368 

101 

679146 

.541 

320854 

28 

33 

634778 

440 

955307 

101 

679471 

641 

320529 

2"/ 

34 

635042 

440 

95.5247 

101 

679795 

541 

320205 

26 

15 

635306 

439 

9.55186 

101 

680120 

640 

319880 

25 

3*6 

635570 

439 

955126 

101 

680444 

640 

319.556 

24 

37 

635834 

439 

95.5065 

lOl 

680768 

640 

319232 

23 

38 

636097 

438 

955005 

101 

081092 

640 

318908 

22 

39 

636300 

438 

954944 

101 

681416 

539 

318.584 

21 

40  J 

636623 

438 

954883 

101 

681740 

52" 

318260; 20  1 

41 

9.638886 

437 

9  9.54823 

101 

9.682063 

539 

10.317937 

19 

42 

637148 

437 

9.54762 

101 

682387 

639 

317613 

18 

43 

637411 

437 

9.54701 

101 

682710 

638 

317290 

17 

44 

637673 

437 

954640 

101 

683033 

538 

316967 

16 

45 

637935 

436 

954579 

101 

683356 

6.38 

316644 

16 

46 

638197 

436 

954518 

102 

683679 

638 

316321 

14 

47 

638458 

436 

954457 

102 

684001 

637 

31.5999 

13 

48 

638720 

435 

9.54396 

102 

684324 

537 

31.5676 

12 

-i9 

638981 

435 

954335 

102 

684646 

637 

315.354 

11 

50 

639242 

435 

954274 

102 

684968 

637 

316032 

10 

51 

9. 639503 

434 

0.954213 

102 

9.685290 

.536 

10.314710 

9 

52 

639764 

434 

954152 

102 

68.56:2 

536 

314388 

8 

53 

640024 

434 

954090 

102 

685934 

536 

314066 

7 

54 

640284 

433 

954029 

102 

686255 

636 

313745 

6 

55 

C40544 

433 

953968 

102 

686577 

C35 

313423 

6 

56 

640804 

433 

9o3906 

102 

686898 

635 

3I3H2 

4 

.'iT 

641064 

432 

953845 

102 

687219 

535 

312781 

3 

58 

641324 

432 

953783 

102 

687540 

635 

312460 

2 

59 

641584 

432 

953722 

103 

687861 

634 

312139 

1 

60 

641842 

431 

9536601 103 

688182 

534 

311818 

-J> 

EJ 


Colatig. 


Taiijr. 


64  l)em«e8. 


44 

(2 

G  Deprr 

eos.)  A 

TABLE  OF  LOGATlirTTMlC 

M. 

Si.u. 

1.. 

C-ine 

1  '>■ 

r.  ... 

i) 

1   Coi.ing.   )   1 

~"o 

!j,H-liy.iv 

431 

9.953660 

io;i 

9.CS8182 

534 

10.311818 

60 

1 

642101 

431 

953599 

103 

688502 

534 

311498 

59 

2 

64-^300 

431 

b53537 

103 

688823 

534 

311177 

58 

3 

642618 

41) 

953475 

103 

689143 

533 

310857 

57 

4 

642877 

430 

9.53413 

103 

6Sii463 

533 

310.537 

56 

5 

643135 

430 

95.3352 

103 

689783 

533 

310217 

55 

6 

643393 

430 

953290 

103 

690103 

533 

309807 

54 

7 

643650 

429 

953228 

103 

690423 

533 

309577 

53 

8 

643908 

429 

953166 

103 

690742 

632 

309258 

52 

9 

644165 

429 

9.53104 

103 

691062 

532 

308938 

51 

10 

644423 

428 

953042 

103 

691.381 

532 

308619 

50 

11 

9.644680 

428 

9.9.52980 

104 

9.H91700 

531 

10.308300 

49 

12 

644936 

428 

952918 

,104 

692019 

531 

307981 

48 

13 

645193 

427 

952855 

104 

692338 

531 

307662 

47 

14 

645450 

427 

952793 

104 

692656 

531 

307344 

46 

15 

645706 

427 

9.52731 

104 

692975 

531 

307025 

45 

16 

645962 

426 

9.52-669 

104 

693293 

530 

306707 

44 

17 

646218 

426 

9.52606 

104 

693612 

530 

30638« 

43 

18 

646474 

426 

9.52.544 

104 

693930 

530 

306070 

42 

19 

646729 

425 

952481 

104 

694248 

530 

305752 

41 

20 
21 

646984 
9.617240 

425 

9.52419  104 
9.9523.56  104 

694566 

529 

30.5434 
10.305117 

40 

39 

425 

9.694883 

.529 

22 

647494 

424 

952294 

104 

69.5201 

529 

304799 

38 

23 

647749 

424 

952231 

104 

69.5518 

529 

304482 

37 

24 

648004 

424 

9.52168 

105 

695836 

529 

304164 

36 

9J} 

648258 

424 

9.V2106 

105 

696153 

528 

303847 

35 

26 

648512 

423 

952043 

105 

696470 

528 

303530 

34 

27 

648766 

423 

951980 

105 

696787 

528 

30.3213 

33 

2vS 

649020 

423 

951917 

105 

697103 

528 

302897 

3-^ 

29 

649274 

422 

9518.54 

105 

697420 

527 

302.580 

31 

30 
31 

649527 

422 

951791 

105 
10.5 

697736 
9.6980.53 

.527 
527 

302264 
10.301947 

30 
29 

9.649781 

422 

9.951728 

32 

6.50034 

422 

951065 

105 

698369 

527 

301631 

28 

3.3 

650287 

421 

951602 

105 

698685 

.526 

301315 

27 

34 

650539 

421 

951.539 

105 

699001 

.526 

300999 

26 

35 

650792 

421 

951476 

105 

699316 

526 

300684 

25 

30 

651044 

420 

951412 

105 

699632 

526 

300368 

24 

37 

651297 

420 

951349 

106 

699947 

526 

300053 

23 

38 

651.549 

420 

951286 

J  06 

700263 

525 

299737 

22 

39 

651800 

419 

951222 

106 

700578 

525 

299422 

21 

40 
41 

652052 
9.6.52.304 

419 
419 

9511.59 

106 
106 

700893 

.525 

299107 
10.298792 

20 
19 

9.951096 

9.701208 

524 

42 

652555 

418 

951032 

106 

701.523 

524 

298477 

18 

43 

6.52806 

418 

950968 

106 

701837 

524 

298163 

17 

44 

653057 

418 

950905 

106 

702152 

524 

297848 

16 

45 

653308 

418 

9.50841 

108 

702466 

524 

297534 

15 

46 

'  653558 

417 

950778 

106 

702780 

523 

297220 

14 

47 

653808 

417 

9.50714 

106 

703095 

523 

296905 

13 

48 

654059 

417 

950650 

106 

703409 

523 

296.591 

12 

49 

654309 

416 

950586 

106 

703723 

523 

296277 

11 

50 

6.54558 

416 

950522 

107 

704036 

522 

295964 

10 

51 

9.6.54808 

416 

9.9504.58 

107 

9.7043.50 

622 

10.29.56.50 

9 

52 

6550.58 

416 

9.50394 

107 

704663 

522 

295337 

8 

53 

65.5.307 

415 

950330 

107 

704977 

522 

29.5023 

7 

54 

655.556 

415 

9.50266 

107 

70.5290 

522 

,  294710 

6 

55 

655805 

415 

950202 

107 

705603 

521 

294397 

5 

56 

6560,54 

414 

9.50138 

107 

705916 

521 

204084 

4 

57 

656302 

414 

950074 

107 

706228 

.521 

293772 

3 

58 

65657)1 

414 

9.50010 

107 

7065 il 

621 

293459 

2 

59 

656799 

413 

949945 

107 

70685 1 

.521 

293 146 

1 

60 

657047 

413 

949881 

107 

707166 

520 

202834 

0 

entitle 

1 

Si.e   1 

(*ii  anir. 

1   '••"•"■'•   1  ''■  1 

()3  Degrees. 


SINES  ATiD  TA.NGE1VTS.  (27  Begi'eei 

'0 

45 

]\j_ 

sn.   1 

^> 

Cosine   1  I). 

Tiintr. 

D- 

1  C'dtaiu'..   1 

~v 

? . 657047 

413 

9.949881 

107 

9.707166 

520 

l0.2J-^«:>4|.i., 

1 

657295 

413 

949816 

107 

707478 

520 

292522  !,Q 

2 

657542 

412 

949752 

107 

707790 

520 

292210  f^ 

3 

657790 

412 

949688 

108 

708102 

520 

291898  57 

4 

tM:8037 

412 

949623 

108 

708414 

519 

291.586  .56 

5 

658284 

412 

9495.58 

108 

708726 

519 

291274 

55 

n 

658531 

411 

949494 

108 

709037 

519 

290963 

54 

7 

658778 

411 

949429 

108 

709349 

519 

290651 

53 

8 

659025 

411 

949364 

108 

709660 

519 

290340 

52 

9 

659271 

410 

949300 

108 

709971 

518 

290029 

51 

10 

659517 

410 

949235 

108 

710282 

518 

289718 

50 

11 

9.659763 

410 

9.949170 

108 

9.710593 

518 

10.289407 

49 

12 

660009 

409 

949105 

108 

710904 

518 

289096 

48 

13 

660255 

409 

949040 

108 

711215 

518 

288785 

47 

14 

660501 

409 

948975 

108 

711.525 

517 

288475 

46 

15 

660746 

409 

948910 

108 

711836 

517 

288164 

45 

16 

660991 

408 

948845 

108 

712146 

517 

287854 

44 

17 

661236 

408. 

948780 

109 

712456 

517 

287.5'U 

43 

18 

661481 

408 

948715 

109 

712766 

516 

287234 

42 

19 

661726 

407 

948650 

109 

713076 

516 

286924 

41 

20 

661970 

407 

948584 

109 

713386 

516 

2866 14 

40 

21 

9.662214 

407 

9.948519 

109 

9.713696 

516 

10.286304 

39 

22 

662459 

407 

948454 

109 

714005 

516 

285995 

38 

23 

662703 

406 

948388 

109 

714314 

515 

285686 

37 

24 

662946 

406 

948323 

109 

714624 

6:5 

285376 

36 

25 

663190 

406 

948257 

109 

714933 

5W 

285067 

35 

26 

663433 

405 

948192 

109 

71.5242 

515 

284758 

34 

27 

663677 

405 

948126 

109 

715551 

514 

£84449 

33 

28 

663920 

405 

948060 

109 

715860 

514 

284140 

32 

29 

664163 

405 

947995 

110 

716168 

514 

283832 

31 

30 
3l' 

664406 

404 

947929 

110 
110 

716477 
9.716785 

514 

283523 
10.28.3215 

30 
29 

9.664648 

404 

9.947863 

514 

32 

664891 

404 

947797 

110 

717093 

513 

282907 

28 

33 

665133 

403 

947731 

110 

717401 

513 

282599 

27 

34 

665375 

403 

947665 

110 

717709 

613 

2S229 1 

26 

35 

665617 

403 

947600 

110 

718017 

513 

281983 

?.5 

36 

665859 

402 

947533 

110 

718325 

•518 

28167.J 

24 

37 

666100 

402 

947467 

110 

718633 

512 

281367 

23 

38 

606342 

402 

947401 

110 

718940 

512 

281000 

22 

39 

666583 

402 

947335 

110 

719248 

512 

280752 

21 

40 
41 

666824 

401 

947269 

110 
110 

719.555 

512 
512 

280445 

20 
19 

9.667065 

401 

9.947203 

9.719862 

10.280138 

42 

667305 

401 

947136 

111 

720169 

511 

279831 

18 

43 

667546 

401 

947070 

111 

720476 

511 

279524 

17 

44 

667786 

400 

947001 

111 

720783 

511 

279217 

16 

45 

668027 

400 

946937 

ni 

721089 

511 

278911 

16 

46 

668267 

400 

946871 

111 

721396 

511 

278604 

14 

47 

668506 

399 

946804 

111 

721702 

510 

278298 

13 

48 

668746 

399 

946738 

111 

722009 

510 

277991 

12 

49 

€68986 

399 

946671 

111 

722315 

510 

277685 

11 

50 

51 

669225 
9.689464 

399 
398 

946604 
9.9465.38 

111 
111 

722621 

610 

277379 
10.277073 

10 
9 

9.722927 

510 

52 

669703 

398 

946471 

111 

723232 

509 

276768 

8 

53 

669942 

398 

946404 

111 

723538 

609 

270462 

7 

54 

670181 

397 

946337 

111 

723844 

609 

2761.56 

6 

55 

670419 

397 

946270 

112 

724149 

509 

27.5851 

5 

56 

670658 

397 

946203 

112 

7244.54 

509 

275546 

4 

57 

670896 

397 

946136 

112 

724759 

508 

275241  3| 

58 

671134 

396 

946069 

112 

725065 

608 

274935 

2 

59 

671372 

396 

946002 

112 

725309 

608 

274631 

1 

= 

671609 

396 

945935 

112 

725674 

508 

274326 

0 



C.isiue 

Sine    1 

Cotaii?. 

1    Tawti.  |M.  j 

62  Degrees. 


46  (28  Defrrces.j     a   tablr   of  LonfAKiTiiMTC 


M 

1     Sl:.- 

1  1». 

'   «Ji)i«in(! 

1  I) 

•;    I'ai;-. 

1   ". 

1    »•..;,„. 

1 

1) 

,   9.67lfi0< 

;}96" 

9.915935  m 

r"9.72567l 

H  50^ 

10. 27 13;;  j 

|-6(r 

1 

67 1847 

395 

915SW  11-^ 

7259/91  508 

274021 

59 

f! 

6720S'i 

395 

9458001  ir* 

726S8^ 

507 

27371(1 

58 

672321 

395 

945733!  Hi 

72658fc 

507 

273412 

57 

672558 

395 

945666  112 

72639-^ 

507 

273108 

56 

5 

,672795 

394 

945598  112 

72719? 

507 

2728G3 

55 

6 

"a73032 

394 

945531  :i2 

727501 

507 

272499 

54 

7 

673268 

3.94 

945464  113 

-727805 

506 

272195 

53 

8 

673505 

394 

945396  113 

728109 

506 

271891 

52 

9 

673741 

393 

945328  113 

728412 

506 

271588 

51 

10 

673977 

1   393 

9452(31]  113 

728716 

506 

2712841  50 

U 

9.674213 

i  393 

9.945193 

1  113 

9.729020 

50ii 

10.270980 

49 

12 

674448 

392 

945125 

i  113 

729323 

505 

270677 

48 

13 

674684 

392 

945058 

1113 

729626 

505 

270374 

47 

14 

674919 

392 

9449901 113 

729929 

505 

270071 

46 

15 

675155 

392 

944922! 113 

730233 

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53 

722791 

338 

92S972 

131 

793819 

469 

206181 

7 

54 

722994 

338 

928893 

131 

794101 

469 

205899 

6 

55 

723197 

338 

928815 

131 

794383 

469 

^05617 

5 

56 

723400 

338 

928736 

131 

794664 

469 

20.5336 

4 

57 

723603 

337 

928657 

131 

794945 

469 

205055 

3 

58 

723805 

337 

928578 

131 

795227 

469 

204773 

2 

59 

724007 

337 

928499 

131 

795508 

468 

204492 

1 

fiO 

724210 

337 

928420  131 

795789 

468 

204211 

0 

~l 

(;...Miiu  1 

time   1 

Cn  ail-:. 

1 

■Vnuii.        j  MTj 

5d  Degreed . 


60 

(32  Degi 

•ees.)  A 

TABLE  OF  tOOARITHMIO 

^ 

M 

1     SilU! 

n. 

1   Cosine   1  D. 

1   Taiifi. 

1   D. 

1   Ooian-;.  | 

~0 

ii.l'Z'i-ZiO 

3^7 

u.yiib'i-^u 

132 

9.796789 

468 

10.204211.60 

1 

724412 

337 

928342 

132 

796070 

468 

203930!. 59 

2 

724614 

336 

928263 

132 

796351 

468 

203649 

[58 

3 

724816 

336 

928183 

132 

796632 

468 

203368 

57 

4 

725017 

335 

928104 

132 

796913 

468 

203087 

56 

5 

725219 

336 

928025 

132 

797194 

468 

202806 

55 

6 

725420 

335 

927946 

132 

797475 

468 

202525 

54 

7 

725622 

335 

927867 

132 

797755 

468 

202245 

53 

8 

725823 

335 

927787 

132 

798036 

467 

201964 

52 

9 

726024 

335 

927708 

132 

798316 

467 

201684 

51 

10 
11 

726225 

335 

927629 

132 
132 

798596 
9.798877 

467 
467 

201404 
10.201123 

50 
49 

9.726426 

334 

9.927549 

12 

726626 

334 

927470 

133 

799157 

467 

200843 

48 

13 

726827 

334 

927390 

133 

799437 

467 

200563 

47 

14 

727027 

334 

927310 

133 

799717 

467 

200283 

46 

15 

727228 

334 

927231 

133 

799997 

466 

200003 

45 

16 

727428 

333 

927151 

133 

800277 

466 

199 723 

44 

17 

727628 

333 

927071 

133 

800557 

466 

199443 

43 

18 

727828 

333 

926991 

133 

800836 

466 

199164 

42 

19 

728027 

333 

926911 

133 

801116 

466 

198884 

41 

20 

728227 

333 

926831 

133 

801396 

466 

198604 

40 

21 

9.728427 

332 

9.92675] 

133 

9.801676 

466 

10.198325 

39 

22 

728626 

332 

926671 

133 

801955 

466 

198045 

38 

23 

728825 

332 

926591 

133 

802234 

465 

197766 

37 

24 

729024 

332 

926511 

134 

802513 

465 

197487 

36 

25 

729223 

331 

926431 

134 

802792 

465 

197208 

35 

26 

729422 

.331 

926351 

134 

80,3072 

465 

196928 

34 

27 

729621 

331 

926270 

134 

803351 

465 

196649 

33 

28 

729820 

331 

926190 

134 

803630 

465 

196370 

32 

29 

730018 

330 

926110 

134 

803908 

465 

196092 

131 

30 

730216 

330 

926029 

134 

804187 

465 

19,5813 

30 

31 

9.730415 

330 

9.925949 

134 

9.804466 

464 

10.19.55.34 

29 

32 

730613 

330 

925868 

134 

804745 

464 

19.52.55 

28 

33 

730811 

330 

925788 

134 

80.5023 

464 

194977 

27 

34 

731009 

329 

925707 

134 

805302 

464 

194698 

26 

35 

731206 

329 

92.5026 

134 

805580 

464 

194420 

25 

36 

731404 

329 

925545 

135 

805859 

464 

194141 

24 

37 

731602 

329 

925465 

135 

806137 

464 

193863 

23 

38 

731799 

329 

925384 

135 

806415 

463 

193585 

22 

39 

731996 

328 

92i3303 

135 

806693 

463 

19.3307 

21 

40 
41 

732193 

328 

925222 

135 
135 

806971 
9.807249 

463 

19.3029 

20 
19 

9.732390 

328 

9.925141 

463 

10.192751 

42 

732587 

328 

925060 

135 

807.527 

463 

192473 

18 

43 

732784 

328 

924979 

135 

807805 

463 

192195 

17 

44 

732980 

327 

924897 

1.35 

808083 

463 

191917 

16 

45 

.  733177 

.327 

924816 

135 

808361 

463 

191639  15 

46 

733373 

327 

924735 

136 

808638 

462 

191362  14 

47 

733569 

327 

9246.54 

1,36 

808916 

462 

191084  13 

48 

733765 

327 

924572 

136 

809193 

462 

190807  12 

49 

733961 

326 

924491 

136 

809471 

462 

190529  11 

50 

734157 

326 

924409 

136 

809748 

462 

190252  10 

51 

9.734353 

320 

9.924328 

136 

9.810025 

462 

10.189975  9 

52 

734549 

326 

924246 

136 

810.302 

462 

189698  8 

53 

734744 

325 

924164 

136 

810580 

462 

189420!  7 

54 

7.34939 

325 

924083 

136 

8108.57 

462 

'  189143 

6 

55 

735135 

325 

924001 

136 

811134 

461 

188866 

M 

56 

735330 

325 

923919 

1.36 

811410 

461 

188590 

4 

57 

735525 

325 

92.3837 

136 

811687 

461 

188313 

3 

58 

735719 

324 

923755 

137 

8; 1964 

461 

188036 

2 

59 

735914 

324 

923673 

137 

812241 

461 

1877.'-^9 

1 

60_ 

736109 

324 

92:^591 

137 

815517 

461 

1874 83 

0 

b 

Cusine   | 

! 

Sine   1    i 

Cotaiig.  1 

1 

Tatig.   1  M.  1 

57  Degrees. 


SINKS  AXD  TANQKKT 

5.  (^^3  Degrees.) 

51 

M 

1     Sil!<! 

1   '»• 

i    (>M,.   1  1). 

1   'I'.iMK. 

-_i^_ 

C.ii.it.i;.  \ 

"o" 

T.TJoi"u9 

3z4 

923509 

L^l 

y.8i:;i5i7 

461 

10.187482,60 

.  1 

73f)3J3 

324 

!37 

8r*794 

461 

18/206  .59 

2 

7304 9S 

324 

923427 

137 

813070 

461 

180930  .58 

3 

736692 

323 

923345 

1.17 

813347 

460 

186653  57 

4 

73d88t) 

323 

923263 

137 

813623 

460 

186377  .56 

ft 

737080 

323 

923181 

137 

313899 

460 

186101  .55 

6 

737274 

323 

923098 

137 

81M75 

460 

185825  54 

7 

737467 

323 

923010 

137 

814452 

460 

18,5.548 

53 

8 

7376G1 

322 

922933 

137 

814728 

460 

18.5272 

52 

9 

737855 

322 

922851 

137 

815004 

460 

184996 

51 

10 

738048 

322 

922768 

138 

81.5279 

460 

184721 

50 

>1 

9.733241 

322 

9.922686 

133 

9.81.55.55 

459 

10.184445 

49 

i2 

738434 

322 

922603 

138 

81.5831 

459 

184169 

48 

i3 

733627 

321 

922520 

138 

816107 

459 

183893 

4? 

U 

738820 

321 

922438 

138 

816382 

459 

183618 

46 

15 

739013 

321 

922355 

138 

816658 

459 

183342 

45 

16 

739206 

321 

922272 

138 

816933 

459 

183067 

44 

17 

739393 

321 

922189 

138 

817209 

459 

182791 

43 

18 

739590 

320 

922108 

138 

817484 

459 

182516 

42 

19 

739783 

320 

922023 

138 

817759 

459 

182241 

41 

20 

739975 

320 

921940 

133 

818035 

458 

181965 

40 

21 

9.740167 

320 

9.9218.57 

139 

9.818310 

458 

^0. 181090 

39 

22 

740359 

320 

921774 

139 

818,585 

458 

181415 

38 

23 

740550 

319 

921691 

139 

818860 

458 

181140 

37 

24 

740742 

319 

921607 

139 

819135 

4.58 

180865 

36 

25 

740934 

319 

921524 

139 

819410 

458 

180590 

35 

26 

741125 

319 

921441 

139 

819684 

458 

180316 

34 

27 

7413 16 

319 

9213.57 

139 

8199.59 

4.58 

180041 

33 

2S 

741.508 

318 

921274 

139 

8202,34 

458 

179766 

32 

29 

741699 

318 

921190 

139 

820.508 

457 

179492 

31 

30 
31 

741889 

318 

921107 
9.921023 

139 
139 

820783 
9.821057 

457 

179217 

30 
29 

9.742030 

318 

457 

10.178943 

32 

742271 

318 

920939 

140 

8213.32 

457 

178668 

28 

33 

.  742462 

317 

920856 

140 

821606 

457 

178394 

27 

34 

742652 

317 

920772 

140 

821880 

457 

178120 

26 

35 

742842 

317 

920688 

140 

822154 

457 

177846 

25 

36 

743033 

317 

920604 

140 

822429 

457 

177571 

24 

37 

743223 

317 

920520 

140 

82270:i 

4.57 

177297 

23 

38 

743413 

316 

920436 

140 

822977 

456 

177023 

22 

39 

743602 

316 

920352 

140 

823250 

456 

1767.50 

21 

40 

743792 

316 

920268 

140 

823524 

456 

17C476J  20  1 

41 

9.743982 

316 

9.920184 

140 

9.823798 

456 

10.176202 

19 

42 

744171 

316 

920099 

140 

824072 

4.56 

175928 

18 

43 

7443G1 

315 

920015 

140 

824345 

456 

175655 

17 

44 

744550 

315 

919931 

141 

824619 

456 

175.381 

16 

45 

744739 

315 

919846 

141 

824893 

456 

175107 

15 

46 

744928 

315 

919762 

141 

825166 

456 

174834 

14 

47 

7451} 7 

315 

919677 

141 

82.5439 

455 

174.561 

13 

48 

745305 

314 

919593 

141 

825713 

455 

174287 

12 

49 

745494 

314 

919508 

141 

825986 

455 

174014 

11 

50 

745383 

314 

919424 

14) 

826259 

455 

173741 

10 

51 

9.745871 

314 

9.91.9339 

141 

9.826532 

455 

10.173468 

9 

52 

746059 

314 

919254 

141 

826805 

455 

173195 

8 

53 

746248 

313 

919169 

141 

827078 

4.55 

172922 

7 

54 

746436 

313 

919085 

141 

827351 

455 

172649 

6 

55 

746624 

313 

919000 

141 

827624 

455 

172376 

5 

56 

746812 

313 

918915 

142 

827897 

454 

172103 

4 

57 

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313 

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142 

823170 

454 

171830 

3 

58 

747187 

312 

918745 

-142 

82344S 

.454 

171.5.58 

2 

59 

747374 

312 

9186.59 

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828715 

454 

171285 

1 

60 

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312 

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142 

828987 

4.54 

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19* 


56  D'-aievs. 
G2 


52 

f^34  Degrees/)  a 

TAIILK  OF  LOGARITHMIC 

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1  n. 

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312 

9.918574 

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10.171013  60 

1 

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142 

829200 

454 

170740  59 

2 

747933 

312 

918404 

142 

829532 

454 

170468  58 

3 

748123 

311 

918318 

142 

829805 

454 

170195  .57 

1 

74S310 

311 

918233 

142 

830077 

454 

1699231.56 
169051  55 

5 

74S497 

311 

918147 

142 

830349 

453 

fi 

748683 

311 

918002 

142 

830021 

453 

169379  .54 

7 

748  S7i) 

311 

917976 

143 

830893 

453 

109107  53 

8 

749053 

310 

917891 

143 

831165 

453 

16833-)  .52 

9 

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831437 

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10 

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310 

917719 

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831709 

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168291 

50 

11 

9.749015 

310 

9.917631 

143 

9.831931 

453 

loTiosoio 

49 

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749801 

310 

917548 

143 

832253 

453 

167747 

48 

13 

749987 

399 

917462 

143 

832525 

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167175 

47 

14 

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309 

917376 

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832796 

453 

1 67-^04 

46 

15 

750358 

309 

917290 

143 

833068 

4.52 

166932 

45 

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750543 

309 

917204 

143 

833339 

452 

166001 

44 

17 

759729 

309 

917118 

144 

833011 

452 

160389 

43 

18 

750914 

3J3 

917032 

144 

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4.52 

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751099 

308 

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831154 

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308 

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165.575 

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9.834690 

4.52 

10.16.5304 

39 

22 

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144 

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163949 

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27 

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307 

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163678 

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10.162595 

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306 

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91.5646 

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20 

35 

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91.5.5.59 

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305 

91.5472 

145 

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450 

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305 

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91.5210 

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55  DvgroM. 


S1NF.S  A^D  TANGENTS 

.  (.^5  Dccrrees. 

; 

53 

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1   C.  .si  III'   1  l>. 

T,-M„f. 

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y.  758591 

301 

9.913365 

147 

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448 

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I 

758772 

300 

913276 

147 

84.5496 

448 

154.504 

,ir> 

2 

758952 

300 

913187 

148 

845764 

448 

1.54236 

58 

3 

759132 

300 

913099 

148 

846033 

448 

1.53967 

57 

4 

759312 

300 

913010 

148 

846302 

448 

1.53698 

56 

6 

759492 

300 

912922 

148 

846570 

447 

1.5.3430 

55 

6 

759672 

299 

912833 

148 

846839 

447 

153161 

54 

7 

759852 

299 

912744 

148 

847107 

447 

152893 

53 

8 

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299 

9126.55 

148 

847376 

447 

152624 

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9 

760211 

299 

912566 

148 

847644 

447 

1.523.-^6 

51 

10 

760390 

299 

912477 

148 

847913 

447 

1.52087 

50 

li 

9.760569 

298 

9.912388 

148 

9.848181 

447 

10.151819 

49 

12 

760748 

298 

912299 

149 

848449 

447 

151,551 

48 

I  a 

760927 

298 

912210 

149 

848717 

447 

151283 

47 

14 

761106 

298 

912121 

149 

848986 

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151014 

46 

15 

761285 

298 

912031 

149 

8492.54 

447 

1.50746 

45 

16 

761464 

298 

911942 

149 

849522 

447 

150478 

44 

17 

761642 

297 

911853 

149 

849790 

446 

1.50210 

43 

18 

761821 

297 

911763 

149 

8.50058 

446 

149942 

42 

19 

761999 

297 

911674 

149 

850325 

446 

149675 

41 

20 

762177 

297 

911584 

149 

850593 

446 

149407 

40 

21 

9.762356 

297 

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10.5368 153 

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105103i.52 

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12 

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10.101730139 

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23 

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0999141321 

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7941.50 

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10.099136 

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35 

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264 

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40 

21 

9.828439 

231 

9.868670 

192 

9.9.59769 

423 

10.040231 

39 

22 

828578 

231 

8685.55 

192 

960023 

423 

039977 

38 

23 

828716 

231 

868440 

192 

960277 

423 

039723 

37 

24 

8.28855 
8^28993 

230 

868324 

192 

960531 

423 

039469 

36 

25 

230 

868209 

192 

960784 

423 

039216 

35 

20 

829131 

230 

868093 

192 

961033 

423 

038962 

34 

27 

829269 

230 

867978 

193 

961291 

423 

038709 

33 

28 

629407 

230 

867862 

193 

961.545 

423 

038455 

32 

29 

829545 

230 

'  867747 

193 

961799 

423 

038201 

31 

30 

8296S3 

230 

867631 

193 

962052 

423 

037948 

30 

31 

9.829821 

229' 

9.867515 

193 

9.962306 

423 

10.037694 

29 

32 

829959 

229 

867399 

193 

962560 

423 

037440 

28 

33 

830097 

229 

867283 

193 

.962813 

423 

0371871271 

34 

830234 

229 

867167 

193 

963067 

423 

0.36933  261 

35 

830372 

229 

867051 

193 

963320 

423 

036680 

25 

3H 

830509 

229 

866935 

194 

963574 

423 

0.36426 

24 

37 

830646 

229 

866819 

194 

963827 

423 

036173 

23 

38 

830784 

229 

866703 

194 

964081 

423 

035919 

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830921 

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8665861 194 

964335 

423 

035665 

21 

40 

831058 

228 

866470 

194 

964.588 

-  422 

035412 

20 

4! 

9.831195 

228 

9.866353 

194 

9.964S42 

422 

10.0351.58 

19 

42 

831332 

228 

866237 

194 

965095 

422 

034905 

18 

43 

831460 

228 

866120 

194 

965349 

422 

034651 

17 

44 

831606 

228  . 

866004 

195 

965602 

422 

034398 

16 

45 

831742 

228 

8658H7 

195 

96.5855 

422 

034145 

15 

4(5 

831879 

228 

865770 

195 

966109 

422 

0.33891 

14 

47 

832015 

227 

8656.53 

195 

966362 

422 

03363S 

13 

4«S 

832152 

227 

865536 

195 

966616 

422 

033384 

12 

•19 

832288 

227 

86.5419 

195 

96686J 

422 

0.33131 

11 

50 

832425 

227 

865302 

195 

967123 

422 

032877 

10 

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97832561 

227 

9. 865 1 85 

195 

9.967376 

422 

10.032624 

9 

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832697 

227 

865068 

195 

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422 

032371 

8 

53 

832833 

227 

864950 

19.) 

967883 

422 

032117 

7 

54 

832969 

226 

864833 

196 

968136 

422 

,031864 

6 

5; 

833105 

226 

864716  196 

968389 

422 

03 1611 

5 

5f5 

83324 1 

226 

864598 

1 96 

968643 

422 

031357  4| 

57 

833377 

226 

864481 

196 

968896 

422 

031104 

3 

5:S 

833512 

226 

864363 

196 

969149 

422 

030851 

2 

5:) 

833648 

226 

864245 

196 

969403 

422 

030597 

1 

f)0_ 

8337.S3 

226 

864127  196 

969656 

422 

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SINES    AND    TANGftKTS.        (43   DpfrrceS  ) 


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I     r..siii( 


n.l 


I      ». 


0 

I 

2 
3 
4 

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6 

7 

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if 
12 
13 
14 

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17 
18 
19 
20 
21 
22 
23 
24 
25 
26 
27 
28 
29 
30 
31 
32 
33 
34 
35 
30 
37 
3f< 
39 
40 

41 
42 
43 
44 
45 
4G 
4? 
48 
49 
50 
51 
52 
53 
54 
55 
5fi 
57 
58 
59 
60 


0.833783 

226 

9.864127 

1961 

9.969656 

422 

10.030344, 60 

833919 

225 

864010 

196 

969909 

422 

U3009 1 1 59 

831054 

225 

863892 

197 

970162 

422 

029838;  .58 

834189 

225 

863774 

197 

970416 

422 

029584  57 

834325 

225 

863656 

197 

970669 

422 

029331 

50 

834460 

225 

863538 

197 

970922 

422 

029078 

55 

834595 

225 

863419 

197 

971175 

422 

028825 

54 

834730 

225 

863301 

197 

971429 

422 

028571  .53 

834865 

225 

863183 

197 

971682 

422 

028318  52 

834999 

224 

863064 

197 

971935 

422 

028005!  51 

835134 

224 

862946 

198 

972188 

422 

0278121 

50 

9.835269 

224 

9.862827 

198 

9.972441 

422 

10.027559' 

49 

815403 

224 

862709 

19S 

972694 

422 

027306! 

48 

8355.38 

224 

862.590 

198 

972948 

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0270.52 

47 

8.35672 

224 

862471 

198 

973201 

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026799 

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862353 

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973454 

423 

026.546 

45 

83594 1 

224 

862234 

198 

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44 

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802 1 15 

198 

973960 

422 

026040 

43 

836209 

223 

861996 

198 

974213 

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025787 

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223 

861877 

198 

974466 

422 

02.5534 

41 

836477 

233 

8617.58 

199 

974719 

4%2 

02.5281 

40 

9.836611 

223 

9.861638 

199 

9.974973 

423 

10.025027 

39 

836745 

223 

861519 

199 

97.5226 

422 

024774 

38 

836878 

223 

861400 

199 

97.5479 

422 

024.521 

37 

837012 

222 

861280 

199 

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024268 

36 

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861161 

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024015 

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023762 

34 

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222 

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199 

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023509 

33 

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222 

860802 

199 

976744 

422 

023256 

32 

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222 

860682 

200 

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422 

023003 

31 

837812 

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860562 

200 

977250 

422 

022750 

30 

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222 

9.860442 

200 

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422 

10.022497 

29 

838078 

221 

860322 

200 

977756 

422 

022244 

28 

8382 1 1 

221 

860202 

200 

978009 

422 

021991 

27 

838344 

221 

860082 

200 

978262 

422 

021738 

26 

838477 

221 

859962 

200 

978515 

422 

021485 

25 

.  838610 

221 

859842 

200 

978768 

422 

0212.32 

24 

833742 

221 

8.59721 

201 

979021 

422 

020979 

23 

838875 

221 

859601 

201 

979274 

422 

020726 

22 

839007 

221 

859480 

201 

979527 

422 

L   020473 

21 

8.39140 

220 

859360 

201 

979780 

422 

020220 

20 

9.839272 

220 

9.859239 

201 

9.980033 

422 

10.019967 

19 

839404 

220 

859119 

201 

980286 

422 

019714 

18 

839536 

220 

8.58998 

201 

980538 

422 

019462 

17 

839668 

220 

858877 

201 

980791 

421 

019209 

10 

839800 

220 

858756 

202 

981044 

421 

018956 

15 

83:)932 

220 

8.58635 

202 

981297 

421 

018703 

14 

840064 

219 

858514 

202 

981550 

421 

018450 

13 

840196 

219 

858393 

202 

98 1803 

421 

018J97 

12 

840328 

219 

858272 

202 

982056 

421 

017944 

11 

840459 

219 

858151 

202 

982309 

421 

017691 

10 

9.840591 

219 

9.858029 

202 

9.982562 

421 

10.017438 

9 

840722 

219 

857908 

202 

982814 

421 

017186 

8 

8408.54 

219 

857786 

202 

983067 

421 

016933 

7 

840985 

219 

857665 

203 

983320 

421 

016680 

6 

841116 

218 

857543 

203 

983573 

421 

016427 

5 

841247 

218 

857422 

203 

983826 

421 

016174 

4 

841378 

218 

857300 

203 

984079 

421 

015921 

3 

841509 

218 

857178 

203 

9>^4331 

421 

01.5669 

2 

811640 

218 

857056 

203 

984584 

421 

015416 

1 

841771 

218 

856931 

203 

1   984837 

421 

01 51 63 

0 

I     C(..-ii..! 


Col  arm. 


I  nu:,.       |M. 


46  Degrees. 


62 

(14  Degrees^  a 

TABLE  OF  LOGARITHMIC 

>J  1 

Sim..' 

'»•  1 

t'osiiie   1  ji 

T.-.,...    I  I)   , 

Cul.liiy.. 

0 

9.841771 

218 

9.856934 

203 

9.9318371  421 

10.015163 

6U 

I 

841^02 

218 

856312 

203 

935090  421 

014910 

59 

2 

8420:13 

218 

85G690 

204 

935343 

421 

014657 

53 

3 

842163 

217 

856568 

204 

935^96 

421 

014404 

57 

4 

842294 

217 

856446 

204 

985343 

421 

0l415ii 

56 

5 

842424 

217 

856323 

204 

986101 

421 

013S99 

55 

6 

842555 

217 

85320 1 

204 

986354 

421 

013646 

54 

? 

842685 

217 

856078 

204 

986607 

421 

013393 

53 

8 

842S15 

217 

855956 

204 

986860 

421 

013140 

52 

9 

842946 

217 

855333 

204 

937112 

421 

012833 

51 

10 

843076 

217 

855711 

2^ 

937365 

421 

012635 

50 

11 

9.813206 

2i6 

9.855538 

205 

9.987618 

421 

10.012332 

49 

12 

843336 

216 

855465 

205 

987871 

421 

012129 

43 

13 

843466 

216 

855342 

205 

988123 

421 

011877 

47 

14 

843595 

216 

855219 

205 

938376 

421 

011624 

46 

15 

843725 

216 

855096 

205 

988629 

421 

011371 

45 

16 

843S55 

216 

854973 

205 

988882 

421 

011118 

44 

17 

843934 

216 

854850 

205 

989134 

421 

010366 

43 

18 

844114 

215 

854727 

206 

939387 

421 

010613 

42 

19 

844243 

215 

854603 

206 

989640 

421 

010360 

41 

20 

844372 

215 

85+180 

206 

939393 

421 

010107 

40 

21 

9.844502 

2j5 

9.854356 

206 

9. 99 J 145 

421 

10.009355 

3  J 

22 

844631 

215 

854233 

206 

990393 

421 

009602 

33 

23 

841760 

215 

854109 

2(r, 

990651 

421 

009349 

37 

24 

8448S9 

215 

8.53936 

206 

990903 

421 

009007 

36 

25 

845018 

215 

853862 

206 

991156 

421 

008344 

35 

26 

845147 

215 

853738 

206 

991409 

421 

00859 1 

34 

27 

845276 

214 

853614 

207 

991662 

421 

003333 

33 

28 

845405 

214 

853490 

207 

991914 

421 

003036 

32 

23 

845533 

214 

853366 

207 

992167 

421 

007833 

31 

30 

845662 

214 

853242 

207 

992420 

421 

007530 

30 

31 

9.845790 

214 

9.8.53118 

207 

9.992672 

421 

10  007323 

29 

32 

845919 

214 

852994 

207 

992925 

421 

007075 

23 

33 

846047 

214 

852869 

207 

993178 

421 

006S2->. 

27 

34 

846175 

214 

852745 

207 

993430 

421 

006570 

26 

35 

846304 

214 

852620 

207 

993633 

421 

006317 

25 

36 

846432 

213 

852496 

203 

993938 

421 

006064 

24 

37 

846560 

213 

852371 

203 

994189 

421 

005311 

23 

33 

8466S8 

213 

852247 

203 

994441 

42  i 

005559 

22 

39 

840316 

213 

852122 

203 

994694 

421 

0053061  21  1 

40 

846944 

213 

851997 

203 

994947 

421 

005053 

20 

41 

9.847071 

213 

9.851872 

203 

9.9951991  421 

10.004301 

19 

42 

847199 

213 

851747 

203 

995452 

421 

004548 

18 

43 

847327 

213 

851622 

208 

995705 

421 

004295 

17 

44 

84X}5;i 

212 

851497 

209 

995957 

421 

O04043 

16 

45 

8  7583 

212 

851372 

209 

996210 

421 

003790 

15 

46 

•  770} 

212 

851246 

209 

996463 

421 

003537 

14 

47 

847836 

212 

851121 

209 

995715 

4-^1 

0032S5 

13 

48 

847964 

212 

850996 

209 

996963 

421 

003032 

12 

49 

818091 

212 

850870 

209 

997221 

421 

002779 

IL 

50 

848218 

212 

850745 

209 

997473 

421 

002527 

10 

51 

9.848345 

212 

9.850519 

209 

9.997726  421 

10.002274 

9 

52 

848472 

211 

85049!J 

210 

997979  421 

002021 

8 

53 

848599 

211 

850368 

210 

998231  421 

001769 

7 

54 

848726 

211 

850242 

210 

998484  421 

O0i5l6 

6 

55 

848852 

211 

850116 

210 

9987371  421 

001263 

5 

56 

848979 

211 

849990 

210 

998989  421 

OOlOll 

4 

57 

849106 

211 

849364 

210 

999242  421 

0007 5S 

3 

58 

84)232 

211 

849738 

210 

9994951  421 

0005(15 

2 

59 

849359 

211 

849611 

210 

999743 1  421 

000253 

1 

60 

819185 

211 

8494S51210 

10. 000000 1  421 

000000    0 

z 

(Jo:illie 

1   »in.e   1 

1   (;.i  ail-.   1 

Taujj.   |W. 

<5J/f«i;« 


%^S^  ifcJ^*-" 


^  ( 


4 


THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 

AN  INITIAL  FINE  OF  25  CENTS 

WILL  BE  ASSESSED   FOR   FAILURE  TO   RETURN 
THIS   BOOK   ON    THE   DATE   DUE.    THE   PENALTY 
WILL  INCREASE  TO  SO  CENTS  ON  THE  FOURTH 
DAY    AND    TO    $1.00    ON     THE    SEVENTH     DAY 
OVERDUE. 

\    APH    y   J  936 

..     :'50^6 

fi'JG  '^Q  1070      m 

I 

^"^  u     .  ij    IJ/y        ^ 

^  M^^-  AUG  12   m 

3 

1 

LD  21-100m-7,'33 

■'meen^- 


nvm 


// 


^J 


*      -'Ji. 


I 


